NCERT Exemplar Class 10 Maths Solutions Chapter 8 Introduction To Trigonometry And Its Equations

# NCERT Exemplar Class 10 Maths Solutions Chapter 8 Introduction To Trigonometry And Its Equations

Edited By Ravindra Pindel | Updated on Sep 05, 2022 05:16 PM IST | #CBSE Class 10th

NCERT exemplar Class 10 Maths solutions chapter 8- Introductions to Trigonometry and Its Equations- NCERT exemplar Class 10 Maths solutions chapter 8 introduces some ratios of sides in a right-angle triangle, which are called Trigonometric ratios. The learnings of this chapter help solve complex problems at later stages in life, whether it be in engineering or architecture. The NCERT exemplar Class 10 Maths chapter 8 solutions follow an in-depth approach and provide solutions at a comprehensive level, enabling the students to study the NCERT Class 10 Maths effectively.

These Class 10 Maths NCERT exemplar chapter 8 solutions, due to their thorough approach, helps the students to build vital concepts of trigonometry and its equations. The NCERT exemplar Class 10 Maths solutions chapter 8 follow the CBSE Syllabus for Class 10 and covers all the topics prescribed by CBSE.

Apply to Aakash iACST Scholarship Test 2024

##### Applications for Admissions are open.

Question:1

If cos A =4/5 , then the value of tan A is

1. $\frac{3}{5}$ (b) $\frac{3}{4}$ (c) $\frac{4}{3}$ (d) $\frac{5}{3}$

Solution. It is given that cos A = 4/5
$\text{We know that cos}\theta=\frac{Base}{Hypotenuse}$
$\therefore$ value of base = 4
Hypotenuse = 5

Use Pythagoras theorem in $\bigtriangleup$ABC
(Hypotenuse)2 = (Base)2 + (perpendicular)
$\left ( AC \right )^{2}= \left ( BC \right )^{2}+\left ( AB \right )^{2}$
$\left ( 5 \right )^{2}= \left ( 4 \right )^{2}+\left ( AB \right )^{2}$
$25-16= \left ( AB \right )^{2}$
$9= \left ( AB \right )^{2}$
$\sqrt{9}= AB$
$3= AB$
That is value of perpendicular is 3
$\text{Also we know that tan}\theta=\frac{perpendicular}{base}$
$\therefore \tan \, A= \frac{3}{4}$
Hence option (B) is correct.

Question:2

If sin A =1/2 then the value of cot A is
a) $\sqrt{3}$ (b) $\frac{1}{\sqrt{3}}$ (c) $\frac{\sqrt{3}}{2}$ (d) 1

$\\sinA=\frac{1}{2}\text{ for angle }30^0\\cosA=\frac{\sqrt{3}}{2}\\cotA=\frac{cosA}{sinA}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$
Hence option (a) is correct.

Question:3

The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is
(A) – 1 (B) 0 (C) 1 (D)$\frac{3}{2}$

Solution. Given expression is :
[cosec(75° + q) – sec (15° – q) – tan (55° + q) + cot (35° – q)]
[cosec(90 – (15 – q)) – sec (15° – q) – tan (90 – (35 – q)) + cot (35° – q)]
{$\because$ we can write (75 + q) = (90 – (15 – q)) and (55 + q) = (90 – (35 – q))}
{$\because$ cosec (90 – q) = sec q and tan (90 – q) = cot q }
[sec(15 – q) – sec (15 – q) – cot (35 – q) + cot (35 – q)]=0
Hence option (B) is correct

Question:4

Given that sinθ =a/b , then cosθ is equal to
(A) $\frac{b}{\sqrt{b^{2}-a^{2}}}$ (B) $\frac{b}{a}$ (C) $\frac{\sqrt{b^{2}-a^{2}}}{b}$ (D)$\frac{a}{\sqrt{b^{2}-a^{2}}}$

Solution. It is given that sin$\theta$ = a/b
$cos\theta=\sqrt{1-sin^2\theta}=\sqrt{1-\frac{a^2}{b^2}}$

$\therefore \cos \theta = \frac{\sqrt{b^{2}-a^{2}}}{b}$
Hence option (C) is correct.

Question:5

If cos (α + β) = 0, then sin (α – β) can be reduced to
(A) cos β (B) cos 2β (C) sin α (D) sin 2α

$\\\text{Given that }\cos(\alpha+\beta)=0\\\Rightarrow (\alpha+\beta)=90^0...........(1)\\\text{wehave to find }\sin(\alpha-\beta)\\\text{from(1) we can write }\alpha=90-\beta\\\therefore \ \sin(\alpha-\beta)=\sin(90-\beta-\beta)\\=\sin(90-2\beta)=\cos2\beta \\(\text{since sin(90-x)=cosx})$
Hence option B is correct.

Question:6

The value of (tan1° tan2° tan3° ... tan89°) is
(A) 0 (B) 1 (C) 2 (D)1/2

Solution. Given :-tan1° tan2° tan3° ... tan89°
tan1° tan2° tan3° ... tan89°tan87° tan 88° tan89° …(1)
We can also write equation (1) in the form of
[tan (900 – 890) . tan (900 – 880). tan (900 – 870) …… tan 87° . tan 88° tan 89°]
[$\because$ we can write tan 1° in the form of tan (900 – 890) similarly we can write other values]
[cot 890 . cot 880. cot 870 …. tan 870 . tan 880 . tan 890]
$\because$ [tan (902$\theta$) = cot$\theta$ ]
Also
$\left [ \frac{1}{\tan 89^{\circ}}\frac{1}{\tan 88^{\circ}} \frac{1}{\tan 87^{\circ}}\cdots \tan 87^{\circ}.\tan 88^{\circ}\tan 89^{\circ} \right ]$
$\because$ Throughout all terms are cancelled by each other and remaining will be tan45
Hence the value is 1
$\because$ option B is correct.

Question:7

If cos 9α = sinα and 9α < 90° , then the value of tan5α is
(A) $\frac{1}{\sqrt{3}}$ (B) ${\sqrt{3}$ (C) 1 (D) 0

Solution. Given :- cos 9$\alpha$ = sin$\alpha$
cos9 $\alpha$ = cos(90 – $\alpha$)
$\because$ (cos (90 – $\alpha$) = sin$\alpha$)
9$\alpha$ = 90 – $\alpha$
9$\alpha$+ $\alpha$ = 90
10 $\alpha$= 90
$\alpha = \frac{90}{10}$
Now tan 5$\alpha$ is
Put $\alpha$ = 9 we get
tan 5$\times$ (9)
tan 450
= 1
{$\because$ from the table of trigonometric ratios of angles we know that tan 450 = 1}
Hence option C is correct.

Question:8

If ΔABC is right angled at C, then the value of cos (A+B) is
$(A) 0 \ \ \ \ \ (B) 1 \ \ \ \ \ (C)1/2 \ \ \ \ \ (D)\frac{\sqrt{3}}{2}$

Solution. It is given that $\angle$C = 90°

In $\bigtriangleup$ABC
$\angle$A +$\angle$B +$\angle$C = 180 [$\because$ sum of interior angles of triangle is 180°]
$\angle$A + $\angle$B + 90o = 180 [$\because$ C = 90° (given)]
$\angle$A + $\angle$B = 1800 – 900
$\angle$A + $\angle$B = 90° …(1)
cos($\angle$A +$\angle$B) = cos (90°)
cos(90°) = 0
[$\because$ from the table of trigonometric ratios of angles we know that cos 90° = 0]
Hence option A is correct.

Question:9

If sinA + sin2A = 1, then the value of the expression (cos2A+ cos4A) is
(A) 1 (B)1/2 (C) 2 (D) 3

Solution. It is given that sinA + sin2A = 1 …(*)
sinA = 1 – sin2A
sinA = cos2A …(1) ( $\because$ 1 – sin2A = cos2A)
Squaring both sides we get
sin2 A = cos4A …(2)
Hence cos2A + cos4A =
= sinA + sin2A {using (1) and (2)}
= sinA + sin2A = 1 (given)
Hence option (A) is correct.

Question:10

Given that sinα =1/2 and cosβ =1/2 , then the value of (α + β) is
(A) 0° (B) 30° (C) 60° (D) 90°

Solution.
$\\\sin \alpha=1/2\Rightarrow \alpha=30^0\\\cos \beta=1/2\Rightarrow \beta=60^0\\\therefore \alpha+\beta=90^0$
Hence option (D) is correct.

Question:11

Solution.

$\left [ \frac{\sin ^{2}22^{\circ}+\sin ^{2}68^{\circ}}{\cos ^{2}22^{\circ}+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos ^{2}63^{\circ}\sin 27^{\circ} \right ]$
$\frac{\sin ^{2}22^{\circ}+\sin ^{2}68^{\circ}}{\cos ^{2}22^{\circ}+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos ^{2}63^{\circ}\sin \left ( 90^{\circ}-63^{\circ} \right )$
$\left [ \because \sin \left ( 90^{\circ}-\theta \right )= \cos \theta \right ]$
$\frac{\sin ^{2}22^{\circ}+\sin ^{2}68^{\circ}}{\cos ^{2}22^{\circ}+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos 63^{\circ}\times \cos 63^{\circ}$
$= \frac{\sin ^{2}22^{\circ}+\sin ^{2}\left (90^{\circ}-22^{\circ} \right )}{\cos ^{2}\left ( 90^{\circ}-68^{\circ} \right )+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos ^{2}63^{\circ}$
$\begin{Bmatrix} \because 68^{\circ}= \left ( 90^{\circ}-22^{\circ} \right ) & \\ 22^{\circ}= \left ( 90^{\circ}-68^{\circ} \right )& \end{Bmatrix}$
$= \frac{\sin ^{2}22^{\circ}+\cos ^{2}22^{\circ}}{\sin ^{2}68^{\circ}+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos ^{2}63^{\circ}$
$\begin{Bmatrix} \because \sin \left ( 90^{\circ}-\theta \right ) = \cos \theta & \\ \cos \left ( 90^{\circ} -\theta \right )= \sin \theta & \end{Bmatrix}$
$= \frac{1}{1}+1\; \left [ \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right ]$
= 1+ 1 =2
Hence option (B) is correct.

Question:12

If $4 \tan\theta= 3$ then $\left ( \frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta } \right )$ is equal to
$(A)\frac{2}{3}$ $(B)\frac{1}{3}$ $(C)\frac{1}{2}$ $(D)\frac{3}{4}$

Hence option (C) is correct.
$\\\left ( \frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta } \right )\text{ Divide numerator and denominator by and rewriting the given expression}\\\\\Rightarrow ( \frac{4\sin \theta -\cos \theta }{4\sin \theta +\cos \theta })=\frac{4\frac{\sin\theta}{cos\theta}-\frac{cos\theta}{cos\theta}}{4\frac{\sin\theta}{cos\theta}+\frac{cos\theta}{cos\theta}}\\\\=\frac{4\tan\theta-1}{4\tan\theta+1}=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2}\ (\text{Given }4\tan\theta=3)$

Question:13

If sinθ – cosθ = 0, then the value of (sin4θ + cos4θ) is
(A) 1 (B)3/4 (C) 1/2 (D) 1/4

$\sin\theta-\cos\theta=0$
squaring both sides we get
$(\sin\theta-\cos\theta)^2=0$
$\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta=0$
($\therefore$ (a – b)2 = a2 + b2 – 2ab)
$\sin^2\theta+\cos^2\theta=2\sin\theta\cos\theta.......(1)$
$1=2\sin\theta\cos\theta\because (\sin^2\theta+\cos^2\theta=1)$
$\frac{1}{2}=\sin \theta \cos \theta$
Squaring both sides we get
$\frac{1}{4}=\sin^{2} \theta \cos^{2} \theta$ …(2)
Now squaring both side of equation (1) we get
$\left ( \sin ^{2}\theta +\cos ^{2}\theta \right )^{2}= \left ( 2\sin \theta \cos \theta \right )^{2}$
$\left ( \sin ^{2}\theta \right )^{2}+\left ( \cos ^{2}\theta \right )^{2}+2\sin ^{2}\theta \cdot \cos ^{2}\theta = 4\sin ^{2}\theta \cos ^{2}\theta$
$\left [ \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right ]$
$\left ( \sin ^{4}\theta \right )+\left ( \cos ^{4}\theta \right )= 4\sin ^{2}\theta \cos ^{2}\theta -2\sin ^{2}\theta \cos ^{2}\theta$
$\sin ^{4}\theta +\cos ^{4}\theta = 2\sin ^{2}\theta \cos ^{2}\theta$
(Use equation (2))
$\sin ^{4}\theta +\cos ^{4}\theta = 2\left ( \frac{1}{4} \right )$
$\sin ^{4}\theta +\cos ^{4}\theta =\frac{1}{2}$
Hence option (C) is correct.

Question:14

sin (45° + θ) – cos (45° – θ) is equal to
(A) 2cosθ (B) 0 (C) 2sinθ (D) 1

Solution. Here
:$\sin \left ( 45^{\circ}+\theta \right )-\cos \left ( 45^{\circ}-\theta \right )$
Sin[90° - (45°- θ)] – cos(45°- θ)
$\left [ \because \left ( 45^{\circ} +\theta \right ) = \left ( 90^{\circ}-\left ( 45-\theta \right ) \right )\right ]$
Cos(45°- θ) – cos(45°- θ) [$\because$ sin (90 – θ) = cosθ]
= 0
Hence option (B) is correct

Question:15

Solution. Given :
height pole = 6 m
Shadow of pole = $2\sqrt{3}m$
Now make figure according to given condition

Let angle of elevation is $\alpha$
$\therefore \tan \alpha = \frac{Perpendicular}{base}$
$\tan \alpha = \frac{6}{2\sqrt{3}}$
$\tan \alpha = \frac{3}{\sqrt{3}}= \sqrt{3}$
$\tan \alpha = \tan 60^{\circ}$ $\left [ \because \tan 60^{\circ}=\sqrt{3} \right ]$
a = 60°
Hence the Sun's elevation is 60°.

Question:1

$\frac{\tan 47^{\circ}}{\cot 43^{\circ}}= 1$
Taking L.H.S.
$\frac{\tan 47^{\circ}}{\cot 43^{\circ}}$
$\frac{\tan \left ( 90^{\circ} -43^{\circ}\right )}{\cot 43^{\circ}}$ $\left ( \because 47= \left ( 90-43 \right ) \right )$
$\frac{\cot 43^{\circ}}{\cot 43^{\circ}}= 1$ $\left ( \because \tan \left ( 90-\theta \right )= \cot \theta \right )$
Hence L.H.S. = R.H.S.
So, the given expression is true.

Question:2

The value of the expression (cos2 23° – sin2 67°) is positive.

Solution. (cos2 23° – sin2 67°)
= $\left ( \cos ^{2}\left ( 90^{\circ}-67^{\circ} \right )-\sin ^{2}67^{\circ} \right )$ $\left ( \because 23^{\circ}= 90^{\circ}-67^{\circ} \right )$
= Sin267° - sin267° (cos(90-θ) = sin θ)
= 0
Hence the value of the expression is neutral
So, the given statement is false.

Question:3

The value of the expression (sin 80° – cos80°) is negative.

Solution. (sin 80° – cos 80°)
We know that from 0 to 90° sin$\theta$ and cos$\theta$ both are positive i.e.
$0< \sin \theta \leq 90^{\circ}$ (always positive)
$0< \cos \theta \leq 90^{\circ}$ (always positive)
At 45° both the values of sin$\theta$ and cos$\theta$ are the same but after 45° to 90° value of sin is greater than the value of cos$\theta$ Hence sin 80° > cos 80°.
If we subtract a smaller term from bigger than the result is positive.
Hence (sin 80° – cos80°) > 0
So, the given statement is false

Question:4

Prove that $\sqrt{\left ( 1-\cos ^{2}\theta \right )\sec ^{2}\theta }= \tan \theta$

L.H.S
$\sqrt{\left ( 1-\cos ^{2}\theta \right )\sec ^{2}\theta } \cdots \left ( 1 \right )$
We know that
$1-\cos ^{2}\theta = \sin ^{2}\theta\text{ in (1)}$
$\sqrt{\sin ^{2}\theta \cdot \sec ^{2}\theta }$
$=\sqrt{\sin ^{2}\theta \times \frac{1}{\cos ^{2}\theta } }$ $\left ( \because \sec \theta = \frac{1}{\cos \theta } \right )$
$=\sqrt{\frac{\sin ^{2}\theta }{\cos ^{2}\theta }}$
$=\sqrt{\tan ^{2}\theta }= \tan \theta$ $\left ( \because \frac{\sin \theta }{\cos \theta }= \tan \theta \right )$
L.H.S. = R.H.S.
Hence the given expression is true.

Question:5

If cosA + cos2A = 1, then sin2A + sin4A = 1.

Given
cosA + cos2A = 1 …(1)
cos A = 1 – cos2A
cosA = sin2A …(2) ($\because$ sin2$\theta$ = 1 – cos2$\theta$)
$\sin ^{2} A+\sin ^{4}A= 1$
L.H.S.
$\sin ^{2} A+\sin ^{4}A$
$\sin ^{2} A+ \left ( \sin ^{2}A \right )^{2}$
$\cos A+\left ( \cos A \right )^{2}$ (from (2))
cosA + cos2A
= 1 (R.H.S.) (from (1))
Hence sin2A + sin4A = 1
So, the given statement is true.

Question:6

(tanθ + 2) (2 tan θ + 1) = 5 tan θ + sec2θ.

(tanθ + 2) (2 tan θ + 1) = 5 tan θ + sec2θ
Taking L.H.S.
(tanθ + 2) (2 tan θ + 1)
tanθ.(2tan θ+1) + 2(2tan θ +1)
$2\tan ^{2}\theta +\tan \theta +4\tan \theta+2$
$2\tan ^{2}\theta +5\tan \theta+2$
$2\left ( \tan ^{2}\theta +1 \right )+5\tan \theta\: \cdots \left ( 1 \right )$
We know that
$\sec ^{2 }\theta-\tan ^{2}\theta= 1$
$\left ( 1+\tan ^{2}\theta= \sec ^{2}\theta \right )$
Put the above value in (1)we get
$2\sec ^{2}\theta +5\tan \theta \neq 5\tan \theta +\sec ^{2}\theta$
L.H.S. $\neq$ R.H.S.
Hence the given expression is false.

Question:7

If the length of the shadow of a tower is increasing, then the angle of elevation of the sun is also increasing.

Solution. Let us take 2 cases.
Case 1:

Case 2 :

$\theta _{1}$ is the angle when length is small and $\theta _{2}$ is the angle when shadow length is increased.
for finding $\theta _{1}$, $\theta _{2}$ find tan $\theta$
$\tan \theta_{1} = \frac{Perpendicular}{Base}= \frac{height\, of\, tower}{length\, of\, shadow}$
$\tan \theta_{2} = \frac{height\, of\, tower}{length\, of\, shadow}$
In both the case height of the tower is the same but in case 2 length of the shadow is increased and if the length of shadow increased value of $\theta _{2}$ decreased.
Hence the given statement is false.

Question:8

If a man standing on a platform 3 metres above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection.

Solution. According to question.

In the figure $\theta _{1}$ is the angle of elevation and $\theta _{2}$ is the angle of depression of the cloud.
For finding $\theta _{1}$find tanq in $\bigtriangleup$ECD
$\tan \theta _{1}= \frac{perpendicular}{Base}= \frac{DC}{EC}= \frac{H}{\iota }$
Find tan$\theta$ in $\bigtriangleup$ECB for $\theta _{2}$
$\tan \theta _{2}= \frac{BC}{EC}= \frac{3}{\iota }$
Here we found that $\theta _{1}$ and $\theta _{2}$ both are different.
Hence the given statement is false.

Question:9

The value of 2sinθ can be a+1/a , where a is a positive number, and a ≠ 1.

Solution. We know that
-1≤ sin θ ≤ 1
Multiply by 2.
-2≤ 2 sin θ ≤ 2
Here we found that value of 2 sin $\theta$ is lies from – 2 to 2.
But if we take a > 0 and a $\neq$ 1 then
$a+\frac{1}{a}> 2$
For example a = 3
3 + 1/3 = 3.33
Hence $a+\frac{1}{a}$ is always greater than 2 in case of positive number except 1
But value of 2 sin $\theta$ is not greater than 2
Hence the given statement is false

Question:10

$\cos \theta = \frac{a^{2}+b^{2}}{2ab}$ where a and b are two distinct numbers such that ab> 0.

We know that
$-1\leq \cos \theta \leq 1$
We also know that
$\left ( a-b \right )^{2}= a^{2}+b^{2}-2ab$
$\text{Since }$ $\left ( a-b \right )^{2}$ $\text{is a square term hence it is always positive }$
$\left ( a-b \right )^{2}> 0$
a2 + b2 – 2ab > 0
$a^{2}+b^{2}> 2ab$
We observe that $a^{2}+b^{2}$ is always greater than 2ab.
Hence,
$\frac{a^{2}+b^{2}}{2ab}> 1$
Because if we divide a big term by small then the result is always greater than 1.
cos$\theta$ is always less than or equal to 1
Hence the given statement is false.

Question:11

The angle of elevation of the top of a tower is 30°. If the height of the tower is doubled, then the angle of elevation of its top will also be doubled.

Solution. According to question

Case: 1
Here BC is the tower.
Let the height of the tower is H and distance AB = a
In $\bigtriangleup$ABC
$\tan\theta=\frac{Perpendicular}{Base}$
$\tan 30^{\circ}= \frac{H}{a}$ $\left ( \because \theta = 30^{\circ} \right )$
$\frac{1}{\sqrt{3}}= \frac{H}{a}\; \cdots \left ( 1 \right )$ $\left ( \because \tan 30^{\circ} = \frac{1}{\sqrt{3}}\right )$
Case :2 When height is doubled

Here ED = a
In $\bigtriangleup DEF$
$\tan \theta = \frac{2H}{a}$
$\tan \theta = \frac{2}{3}$ (from (1))
But $\tan 60^{\circ}= \sqrt{3}$ (If the angle is double)
$\sqrt{3}\neq \frac{2}{\sqrt{3}}$
hence the given statement is false.

Question:12

If the height of a tower and the distance of the point of observation from its foot,both, are increased by 10%, then the angle of elevation of its top remains unchanged.

Solution. According to question

In case-1. Height is H and observation distance is a.
In case-2, both height and observation distance is increased by 10%.
In case -1
$\tan \theta _{1}= \frac{H}{a}$ $\left ( \because \tan \theta = \frac{perpendicular}{Base} \right )$ .....(1)
In case -2
$\tan \theta _{2}= \frac{H+\frac{H}{10}}{a+\frac{a}{10}}$
$= \frac{\frac{11H}{10}}{\frac{11a}{10}}= \frac{11H}{10}\times \frac{10}{11a}= \frac{H}{a}$
$\tan \theta _{2}= \frac{H}{a} \cdots \left ( 2 \right )$
from equation (1) and (2) we observe that $\theta _{1}= \theta _{2}$
Hence the given statement is true.

Question:1

Prove the following
:$\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta}{\sin \theta}= 2\cos ec\theta$

$\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta}{\sin \theta}= 2\cos ec\theta$
Taking L.H.S.
$= \frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta}{\sin \theta}$
Taking LCM
$= \frac{\sin ^{2}\theta +\left ( 1+\cos \theta \right )^{2}}{\left ( 1+\cos \theta \right )\sin \theta }$
$= \frac{\sin ^{2}\theta +\cos ^{2}\theta +2\cos \theta }{\left ( 1+\cos \theta \right )\sin \theta }$ $\left ( \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right )$
$= \frac{1+1+2\cos \theta }{\left ( 1+\cos \theta \right )\sin \theta }$ $\left ( \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right )$
$= \frac{2\left ( 1+\cos \theta \right )}{\left ( 1+\cos \theta \right )\sin \theta }$
$= \frac{2}{\sin \theta }$
$= 2 \cos ec\theta$ $\left ( \because \frac{1}{\sin \theta } = \cos ec \theta \right )$

L.H.S. = R.H.S.
Hence proved.

Question:2

Prove the following :
$\frac{tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}= 2cosecA$

Solution.
$\frac{tan A}{1+\sec A}-\frac{\tan A}{1-\sec A} = 2cosecA$
Taking L.H.S.
$= \frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}$
Taking L.C.M.
$=\frac{\tan A\left ( 1-\sec A \right )-\tan A\left ( 1+\sec A \right )}{\left ( 1+\sec A \right )\left ( 1-\sec A \right )}$
$=\frac{\tan A-\tan A\\sec A-\tan A-\tan A\sec A}{1-\sec ^{2}A}$ $\left ( \because \left ( a-b \right )\left ( a+b \right ) = a^{2}-b^{2}\right )$
$= \frac{-2\tan A\sec A}{-\tan ^{2}A}$ $\left ( \because \sec ^{2}A-\tan ^{2}A= 1 \right )$
$= \frac{2\sec A}{\tan A}$
$= \frac{2}{\cos A}\times \frac{\cos A}{\sin A}$ $\begin{pmatrix} \because \sec \theta = \frac{1}{\cos \theta } & \\ \because \tan \theta =\frac{\sin \theta }{\cos \theta } & \end{pmatrix}$
$= \frac{2}{\sin A}= 2\cos ec A$ $\left ( \because \frac{1}{\sin \theta }= \cos ec\theta \right )$
L.H.S. = R.H.S.
Hence proved

Question:3

If tan A = 3/4 , then show that sinAcos A = 12/25 .

Given:-
$\tan A= \frac{3}{4}$
To prove:-
$sin A\cos A= \frac{12}{25}...........(1)$
we know that
$\tan \theta = \frac{P}{B}$
$\frac{P}{B}= \frac{3}{4}$
P=3, B = 4
Using Pythagoras theorem
$H^{2}= B^{2}+P^{2}$
$H^{2}= \left ( 4^{2} \right )+\left ( 3^{2} \right )$
$H^{2}= 9+16$
$H^{2}= 25$
$H= \sqrt{25}= 5$
H = 5
we know that
$\sin \theta = \frac{P}{H},\cos \theta = \frac{B}{H}$
Hence
$\sin A= \frac{3}{5},\cos A= \frac{4}{5}$
Put the value of sinA and cosA in equation (1)
$\frac{3}{5}\times \frac{4}{5}= \frac{12}{25}$
L.H.S. = R.H.S.
Hence proved.

Question:4

Prove the following : (sin α + cos α) (tan α + cot α) = sec α + cosec α

Solution.
(sin α + cos α) (tan α + cot α) = sec α + cosec α
Taking L.H.S.
$\left ( \sin \alpha +\cos \alpha \right )\left ( \tan \alpha +\cot \alpha \right )$
$= \left ( \sin \alpha +\cos \alpha \right )\left ( \frac{\sin \alpha }{\cos \alpha } +\frac{\cos \alpha}{\sin \alpha } \right )\left ( \because \tan \theta = \frac{\sin \theta }{\cos \theta },\cot \theta = \frac{\cos \theta }{\sin \theta } \right )$
Taking L.C.M.
$= \frac{\left ( \sin \alpha +\cos \alpha \right )\left ( \sin ^{2}\alpha +\cos^{2} \alpha \right )}{\sin \alpha +\cos \alpha }$
$= \frac{\sin \alpha }{\sin \alpha \cos \alpha }+\frac{\cos \alpha }{\sin \alpha +\cos \alpha }$ $\left ( \because \sin^{2} \theta +\cos^{2} \theta = 1 \right )$
by separately divide
$= \frac{\sin \alpha }{\sin \alpha \cos \alpha }+\frac{\cos \alpha }{\sin \alpha +\cos \alpha }$
$= \frac{1}{\cos \alpha }+\frac{1}{\sin \alpha }$
$= \sec \alpha +\cos ec \, \alpha$ $\left ( \because \frac{1}{\cos \theta }= \sec \theta ,\frac{1}{\sin \theta }= \cos ec\theta \right )$
L.H.S. = R.H.S.
Hence proved.

Question:5

$\left ( \sqrt{3}+1 \right )\left ( 3-\cot 30^{\circ} \right )= \tan ^{3}60^{\circ}-2\sin 60^{\circ}$
Taking L.H.S
$\left ( \sqrt{3}+1 \right )\left ( 3-\cot 30^{\circ} \right )$
$= \sqrt{3}\left ( 3-\cot 30^{\circ} \right )+1\left ( 3-\cot 30^{\circ} \right )$
$=3 \sqrt{3}-\sqrt{3}\cot 30^{\circ}+3-\cot 30^{\circ}$
We know that
$\cot 30^{\circ}= \sqrt{3}$
$= 3\sqrt{3}-\sqrt{3}\left (\sqrt{3} \right )+3-\sqrt{3}$
$= 3\sqrt{3}-3+3-\sqrt{3}$
$= 3\sqrt{3}-\sqrt{3}$
$= 2\sqrt{3}$
R.H.S.
$\tan ^{3}60^{\circ}-2\sin 60^{\circ}$
We know that
$tan 60^o = \sqrt{3}$
$\tan ^{3}60^{\circ}-2\sin 60^{\circ}= \left ( \sqrt{3} \right )^{3}-2\left ( \frac{\sqrt{3}}{2} \right )$
$= 3\sqrt{3}-\sqrt{3}= 2\sqrt{3}$
Hence proved

Question:6

Prove the following :
$1+\frac{\cot ^{2}\alpha }{1+\cos ec\alpha }= \cos ec\, \alpha$

$1+\frac{\cot ^{2}\alpha }{1+\cos ec\alpha }= \cos ec\, \alpha$
Taking L.H.S.
$= 1+\frac{\cot ^{2}\alpha }{1+\cos ec\, \alpha }$

$= \frac{1+\cos ec\, \alpha +\cot ^{2}\alpha }{1+\cos ec\, \alpha }$
$= \frac{\cos ec^{2}\alpha-\cot ^{2}\alpha+ \cos ec\, \alpha+\cot ^{2}\alpha \, \, }{1+\cos ec\, \alpha }$ $\left ( \because \cos ec^{2}\theta -\cot ^{2}\theta = 1 \right )$
$= \frac{\cos ec^{2}\alpha +\cos ec\, \alpha }{1+\cos ec\, \alpha }$
$= \frac{\cos ec\, \alpha \left ( \cos ec\, \alpha+1 \right ) }{\left ( 1+\cos ec\, \alpha \right )}$
$= \cos ec\, \alpha$
L.H.S. = R.H.S.
Hence proved.

Question:7

Prove the following: tan θ + tan (90° – θ) = sec θ sec (90° – θ)

Solution.
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
Taking L.H.S.
= tan θ + tan (90° – θ)
= tanθ + cotθ ($\because$ tan (90 – θ) = cot θ)
$=\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }$ $\left ( \because \tan \theta = \frac{\sin \theta }{\cos \theta },\cot \theta = \frac{\cos \theta }{\sin \theta } \right )$
Taking L.C.M.
$\frac{\sin ^{2}\theta +\cos ^{2}\theta }{\sin \theta \cos \theta }$
$= \frac{1}{\cos \theta \cdot \sin \theta }$ $\left ( \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right )$
$= \frac{1}{\cos \theta }\times \frac{1}{\sin \theta }$
$= \sec \theta \times \cos ec\, \theta$ $\left ( \because \frac{1}{\cos \theta } = \sec \theta ,\frac{1}{\sin \theta }= \cos ec\, \theta \right )$
$= \sec \theta \times \sec \left ( 90^{\circ}-\theta \right )$ $\left ( \because \sec \left ( 90-\theta \right ) = \cos ec\, \theta \right )$
L.H.S. = R.H.S.
Hence proved.

Question:8

Find the angle of elevation of the sun when the shadow of a pole h metres high is $\sqrt{3}$ h metres long.

Solution. According to question

Here BC is the height of the pole i.e. h meters and AB is the length of shadow i.e. $\sqrt{3}h$ .
For finding angle q we have to find tanq in $\bigtriangleup$ABC
$\tan \theta = \frac{Perpendicular}{Base}$
$= \tan \theta = \frac{h}{\sqrt{3}h}$
$\theta = 30^{\circ}$ $\left ( \because \tan 30^{\circ}= \frac{1}{\sqrt{3}} \right )$
Hence angle of elevation is 30°.

Question:9

If $\sqrt{3}$ tan θ = 1, then find the value of sin2θ – cos2θ.

Given :
$\sqrt{3}\tan\theta=1$
$\tan \theta = \frac{1}{\sqrt{3}}$
$\theta = 30^{\circ}$ $\left ( \because \tan30^{\circ}= \frac{1}{\sqrt{3}} \right )$
$\sin ^{2}\theta -\cos ^{2}\theta = \sin ^{2}30^{\circ}-\cos ^{2}30^{\circ}$
(Because θ = 300)
$= \left ( \frac{1}{2} \right )^{2}-\left ( \frac{\sqrt{3}}{2} \right )^{2}$ $\begin{bmatrix} \because \sin 30^{\circ}= \frac{1}{2} & \\ \cos 30^{\circ}= \frac{\sqrt{3}}{2}& \end{bmatrix}$
$= \frac{1}{4}-\frac{3}{4}$
Taking L.C.M.
$= \frac{1-3}{4}= \frac{-2}{4}$
$\sin ^{2}\theta -\cos ^{2}\theta = \frac{-1}{2}$

Question:10

A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, find the height of the wall.

Length of ladder = 15 m
The angle between wall and ladder = 60°
Let the height of wall = H

In $\bigtriangleup$ABC $\angle$C = 60°, $\angle$B = 90°
We know that
$\angle$A + $\angle$B + $\angle$C = 180° (Sum of interior angles of a triangle is 180)
$\angle$A + 90 + 60 = 180°
$\angle$A = 30°
In $\bigtriangleup$ABC
$\sin 30^{\circ} = \frac{H}{15}$
$H= \sin 30^{\circ} \times 15$
$H= \frac{1}{2} \times 15$ $\left ( \because \sin 30^{\circ} = \frac{1}{2}\right )$
$H= 7\cdot 5\, m$
Hence the height of the wall is 7.5 m

Question:11

$\text{Simplify} (1+tan^2\theta)(1-sin\theta)(1+sin\theta)$

$(1+tan^2\theta)(1-sin\theta)(1+sin\theta)$
$= \left ( \sec ^{2}\theta \right )\left ( \left ( 1 \right )^{2} -\left ( \sin \theta \right )^{2}\right )$ $\left ( \because \left ( a-b \right )\left ( a+b \right ) = a^{2}-b^{2}\right )$
$= \left ( \sec ^{2}\theta \right )\left ( 1-\sin ^{2}\theta \right )$
$= \left ( \sec ^{2}\theta \right )\left ( \cos ^{2}\theta \right )$ $\left ( \because \sin ^{2}\theta + \cos ^{2}\theta= 1 \right )$
$= \frac{1}{ \cos ^{2}\theta}\times \cos ^{2}\theta$ $\left ( \because \sec ^{2}\theta = \frac{1}{\cos \theta } \right )$
= 1

Question:12

If 2sin2θ – cos2θ = 2, then find the value of θ.

2sin2θ – cos2θ = 2
$2\left ( 1-\cos ^{2}\theta \right )-\cos ^{2}\theta = 2$ $\left ( \because \sin ^{2}\theta +\cos ^{2}\theta= 1 \right )$
$2-2\cos ^{2}\theta-\cos ^{2}\theta= 2$
$2-3\cos ^{2}\theta-2= 0$
$-3\cos ^{2}\theta= 0$
$\cos ^{2}\theta= 0$
$\cos \theta= 0$
$\theta= 90^{\circ}$ $\left ( \because \cos 90^{\circ}= 0 \right )$
Hence value of $\theta$ is 90°

Question:13

Show that
$\frac{\cos ^{2}\left ( 45^{\circ}+\theta \right )+\cos ^{2}\left ( 45^{\circ}-\theta \right )}{\tan \left ( 60^{\circ} +\theta \right )\tan \left ( 30 ^{\circ}+\theta \right )}= 1$

L.H.S
$= \frac{\cos ^{2}\left ( 90-\left ( 45-\theta ^{\circ} \right ) \right )+\cos ^{2}\left ( 45-\theta ^{\circ} \right ) }{\tan \left ( 60^{\circ}+\theta \right )\tan \left ( 30^{\circ}-\theta \right )}$ $\left ( \because \sin \left ( 90^{\circ}-\theta \right ) = \cos \theta \right )$
$= \frac{\sin ^{2}\left ( 45^{\circ}-\theta \right )+\cos ^{2}\left ( 45^{\circ} -\theta\right )}{\tan \left ( 90^{\circ}-\left (30^{\circ} -\theta \right ) \right )\tan \left ( 30^{\circ}-\theta \right )}$ $\left ( \because \tan \left ( 90-\theta \right )= \cot \theta \right )$
$= \frac{\sin ^{2}\left ( 45-\theta \right )+\cos ^{2}\left ( 45-\theta \right )}{\cot \left ( 30^{\circ}-\theta \right )\tan \left ( 30^{\circ}-\theta \right )}$ $\left ( \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right )$
$= \frac{1}{\frac{1}{\tan \left ( 30^{\circ}-\theta \right )}\times \tan \left ( 30-\theta \right )}$ $\left ( \because \cot \theta= \frac{1}{\tan \theta} \right )$
$= \frac{1}{1}=1$
L.H.S. = R.H.S.
Hence proved.

Question:14

An observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.

Solution. According to the question.

In $\bigtriangleup$EDC EC = 20.5 m
DC = 20.5 m
To find angle $\theta$ in $\bigtriangleup$EDC we need to find tan$\theta$.
$\tan \theta = \frac{P}{B}$
$\tan \theta = \frac{EC}{DC}$
$\tan \theta = \frac{20\cdot 5}{20\cdot 5}$
$\tan \theta = 1$
$\theta = 45^{\circ}$ $\left ( \because \tan 45^{\circ}= 1 \right )$
Hence the angle of elevation is 45°.

Question:15

Show that tan4θ + tan2θ = sec4θ – sec2θ.

Taking L.H.S.
tan4θ + tan2θ
(tan2θ) + tan2θ…(1)
We know that sec2θ – tan2θ = 1
Put

$\tan ^{2}\theta = \sec ^{2}\theta -1$ in (1)
$= \left ( \sec ^{2}\theta -1 \right )^{2}+\sec ^{2}\theta -1$
$= \left ( \sec ^{2}\theta \right )^{2}+\left ( 1 \right )^{2}-2\left ( \sec ^{2}\theta \right )\left ( 1 \right )+\sec ^{2}\theta -1$
$\left ( \because \left ( a-b \right )^{2} = a^{2}+b^{2}-2ab\right )$
$= \sec ^{4}\theta +1-2\sec ^{2}\theta +\sec ^{2}\theta -1$
$= \sec ^{4}\theta -\sec ^{2}\theta$
LHS = RHS
Hence proved

Question:1

If cosecθ + cotθ = p, then prove that
$cos\theta=\frac{p^{2}-1}{p^{2}+1}$ .

Given: cosecθ + cotθ = p …(1)

Taking right hand side.
$\frac{p^{2}-1}{p^{2}+1}$
Put value of p from equation (1) we get
$= \frac{\left ( \cos ec\, \theta +\cot \theta \right )^{2}-1}{\left ( \cos ec\, \theta +\cot \theta \right )^{2}+1}$
$= \frac{ \cos ec\,^{2} \theta +\cot^{2} \theta+2 \cos ec\, \theta \cot \theta-1}{ \cos ec\,^{2} \theta +\cot^{2} \theta+2 \cos ec\, \theta \cot \theta+1}$
$\left [ \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab \right ]$
$= \frac{\frac{1}{\sin ^{2}\theta }+\frac{\cos ^{2}\theta }{\sin ^{2}\theta}+\frac{2\cos \theta}{\sin ^{2}\theta}-1}{\frac{1}{\sin ^{2}\theta }+\frac{\cos ^{2}\theta }{\sin ^{2}\theta}+\frac{2\cos \theta}{\sin ^{2}\theta}+1}$
$\left [ \because \cos ec\, \theta = \frac{1}{\sin \theta } ,\cot \theta = \frac{\cos \theta }{\sin \theta }\right ]$
$= \frac{\frac{1+\cos ^{2}\theta +2\cos \theta -\sin ^{2}\theta }{\sin ^{2}\theta}}{\frac{1+\cos ^{2}\theta +2\cos \theta +\sin ^{2}\theta }{\sin ^{2}\theta}}$ [by taking LCM]
$= \frac{1+\cos ^{2}\theta +2\cos \theta -\sin ^{2}\theta }{1+\cos ^{2}\theta +2\cos \theta +\sin ^{2}\theta}$
$= \frac{1-\sin ^{2}\theta +\cos \theta +2\cos \theta }{1+\sin ^{2}\theta +\cos \theta +2\cos\theta}$
$= \frac{\cos ^{2}\theta +\cos ^{2}\theta+2\cos \theta }{1+1+2\cos \theta }$ $\begin{bmatrix} \because 1-\sin ^{2\theta }= \cos ^{2} \theta & \\ \sin ^{2 }+\cos ^{2}\theta = 1 & \end{bmatrix}$
$= \frac{2\cos ^{2}\theta +2\cos \theta }{2+2\cos \theta}$
$= \frac{2\cos \theta \left ( \cos \theta +1 \right )}{2 \left ( \cos \theta +1 \right )}$
$= \frac{2\cos \theta }{2}$
$= \cos \theta$
which is equal to the eft-hand side
Hence proved.

Question:2

Prove that $\sqrt{\sec ^{2}\theta +\cos ec^{2}\theta }= \tan \theta +\cot \theta$

Taking left-hand side
$\sqrt{\sec ^{2}\theta +\cos ec^{2}\theta }$
$= \sqrt{\frac{1}{\cos ^{2}\theta }+\frac{1}{\sin ^{2}\theta }}$ $\left [ \because \sec \theta = \frac{1}{\cos \theta } ,\cos ec= \frac{1}{\sin \theta }\right ]$
$= \sqrt{\frac{\sin ^{2}\theta +\cos ^{2}\theta }{\cos ^{2}\theta \cdot \sin ^{2}\theta }}$
$= \sqrt{\frac{1}{\cos ^{2}\theta\cdot \sin ^{2} \theta }}$ $\left [ \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right ]$
$\frac{1}{\cos \theta \sin \theta }=\frac{sin^2\theta+cos^2\theta}{cos \theta sin\theta}=\frac{sin^2\theta}{cos \theta sin\theta}+\frac{cos^2\theta}{cos \theta sin\theta}\\\\=\frac{sin\theta}{cos\theta}+\frac{cos\theta}{sin\theta}=tan\theta+cot\theta=RHS\\\text{Hence proved}$

Question:3

The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.

Solution.

The angle of elevation of the top of a tower AB from certain point C is 30°
Let observer moves from C to D that is CD = 20m
Now angle of elevation increased by 15° that is 45° on point D
In $\bigtriangleup$ABD
$\tan 45^{\circ}= \frac{AB}{BD}$ $\left [ \because \tan \theta = \frac{Perpendicular}{Base} \right ]$
$1= \frac{AB}{BD}$
BD = AB …(1)
In $\bigtriangleup$ABC
$\tan 30^{\circ}= \frac{AB}{BC}$ $\left [ \because \tan \theta = \frac{Perpendicular}{Base} \right ]$
$\frac{1}{\sqrt{3}}= \frac{AB}{BD+DC}$ $\begin{bmatrix} \because \tan 30= \frac{1}{\sqrt{3}} & \\ BC= BD+DC & \end{bmatrix}$
$\frac{1}{\sqrt{3}}= \frac{AB}{BD+20}$ $\left ( \because DC= 20 \right )$
By cross multiplication we get
$BD+20= \sqrt{3}AB$
Now put the value of BD from equation (1) we have
$AB+20= \sqrt{3}AB$
$20= \sqrt{3}AB-AB$
$20= AB\left ( \sqrt{3}-1 \right )$
$AB= \frac{20}{\sqrt{3}-1}$
$AB= \frac{20}{1\cdot 732-1}= \frac{20}{0\cdot 732}$
AB = 27.322
Hence the height of the tower is 27.322 m

Question:4

If 1 + sin2θ = 3sinθ cosθ, then prove that tanθ = 1 or1/2

Solution. Given : 1 + sin2θ = 3sinθ cosθ
To Prove - tanθ = 1 or 1/2
Dividing both side by sinθ we get
$\frac{1+\sin ^{2}\theta }{\sin ^{2}\theta}= \frac{3\cos \theta }{\sin \theta }$
$\frac{1}{\sin ^{2}\theta}+\frac{\sin ^{2}\theta}{\sin ^{2}\theta}= 3\cot \theta$ $\left ( \because \frac{\cos \theta }{\sin \theta } = \cot \theta \right )$
$\cos ec^{2}\theta +1= 3\cot \theta$ $\left ( \because \frac{1}{\sin ^{2}\theta}= \cos ec^{2}\theta \right )$
$1+\cot ^{2}\theta +1= 3\cot \theta$ $\left ( \because \cos ec^{2}\theta = 1+\cot ^{2} \theta \right )$
$2+\cot ^{2} \theta+3\cot \theta= 0$
$\cot ^{2} \theta+3\cot \theta+2= 0$
$\cot ^{2} \theta-2\cot \theta-\cot \theta +2= 0$
$\cot \theta\left ( \cot \theta-2 \right )-1\left ( cot \theta-2\right )= 0$
$\left ( \cot \theta-2 \right )\left ( \cot \theta-1 \right )= 0$
$\cot \theta = 1,2$
We know that
$\tan \theta = \frac{1}{\cot \theta }$
$\therefore \tan \theta = 1,\frac{1}{2}$
Hence proved.

Question:5

Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.

Solution. Given:- sinθ + 2cosθ = 1
squaring both sides we have
$\left ( \sin \theta +2\cos \theta \right )^{2}= 1^{2}$
$\sin ^{2}\theta +4\cos ^{2}\theta +4\sin \theta \cos \theta = 1$
$\left ( \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right )$
$\sin ^{2}\theta +4\cos ^{2}\theta = 1-4\sin \theta \cos \theta \cdots \left ( 1 \right )$
To prove :
$2\sin \theta -\cos \theta = 2$
Taking the left-hand side
$2\sin \theta -\cos \theta \cdots \left ( 2 \right )$
On squaring equation (2) we get
$\left (2 \sin \theta -\cos \theta \right )^{2}$
$= 4\sin ^{2}\theta +\cos ^{2}\theta -4\sin \theta \cos \theta$$\left ( \because \left ( a-b \right )^{2} = a^{2}+b^{2}-2ab\right )$
$= 3\sin ^{2}\theta +\sin ^{2}\theta+\cos ^{2}\theta -4\sin \theta \cos \theta$
$= 3\sin ^{2}\theta +1-4\sin \theta \cos \theta$ $\left [ \because \sin ^{2} \theta +\cos ^{2} \theta= 1\right ]$
$= 3\sin ^{2} \theta+\sin^{2} \theta+4\cos ^{2} \theta$
[use equation (1)]
$= 4\sin ^{2}\theta +4\cos ^{2}\theta$
$= 4\left ( \sin ^{2}\theta +\cos ^{2}\theta \right )$
= 4
$\left [ \because \sin ^{2} \theta +\cos ^{2} \theta= 1\right ]$
So here we get the value of (2sin$\theta$ – cos$\theta$)2 is 4
$\left ( 2\sin \theta -\cos \theta \right )^{2}= 4$
$2\sin \theta -\cos \theta = \sqrt{4}$
$2\sin \theta -\cos \theta = 2$
Hence proved

Question:6

The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is$\sqrt{st}$

Solution. According to question

Let the height of tower = h
the distance of the first point from its foot = s
the distance of the second point from its foot = t
$\tan \theta = \frac{h}{s}\cdots \left ( 1 \right )$ $\left ( \because \tan \theta = \frac{Perpendicular}{Base} \right )$
$\tan \left ( 90-\theta \right )= \frac{h}{t}$
$\cot \theta = \frac{h}{t}\cdots \left ( 2 \right )$
Multiply equation (1) and (2) we get
$\tan \theta \times \cot \theta= \frac{h}{t}\times \frac{h}{s}$
$\tan \theta \frac{1}{ \tan \theta }= \frac{h^{2}}{st}$ $\left ( \because \cot \theta = \frac{1}{\tan \theta } \right )$
$1= \frac{h^{2}}{st}$
$st= h^{2}$
$h= \sqrt{st}$
Hence proved.

Question:7

The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower.

Solution. According to question

Let the height of tower = h
$\tan 60^{\circ}= \frac{h}{BD}$ $\left [ \because \tan \theta = \frac{Perpendicular}{Base} \right ]$
$\sqrt{3}= \frac{h}{BD}$ $\left [ \because \tan 60^{\circ}= \sqrt{3} \right ]$
$BD= \frac{h}{\sqrt{3}}\cdots \left ( 1 \right )$
$\tan 30^{\circ}= \frac{h}{BC}= \frac{h}{BD+DC}$ $\left [ \because BC= BD+DC \right ]$
$\frac{1}{\sqrt{3}}= \frac{h}{BD+50}$
$BD+50= \sqrt{3h}$
[by cross multiplication]
$\frac{h}{\sqrt{3}}+50= \sqrt{3h}$
[from equation (1)]
$\frac{h}{\sqrt{3}}= \sqrt{3h}-50$
$h= 3h-50\sqrt{3}$
$h= \frac{50\sqrt{3}}{2}$
$h= 25\sqrt{3}m$

Question:8

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are α and β, respectively. Prove that the height of the tower is
$\left ( \frac{h\tan \alpha }{\tan \beta -\tan \alpha } \right )$

According to question

Here h is the height of flagstaff AD.
Let $\l$ is the height of the tower
$\alpha$ and $\beta$ be the angle of elevation of the bottom and the top of the flagstaff.
In $\bigtriangleup$BDC
$\tan \alpha = \frac{\l }{BC}$ $\left [ \because \tan \theta = \frac{Perpendicular}{Base} \right ]$
$BC= \frac{\l }{\tan \alpha }\cdots \left ( 1 \right )$
In $\bigtriangleup$ABC
$\tan \beta = \frac{AB}{BC}$
$\tan \beta = \frac{h+\l }{BC}$
$BC= \frac{h+\l }{\tan \beta }\cdots \left ( 2 \right )$
equate equation (1) and (2) we get
$\frac{\l }{\tan \alpha }=\frac{h+\l }{\tan \beta }$
$\l \tan \beta = h\tan \alpha +\l \tan \alpha$ [by cross multiplication]
$\l \tan \beta -\l \tan \alpha = h\tan \alpha$
$\l \left ( \l \tan \beta - \tan \alpha \right ) = h\tan \alpha$
$\l = \frac{h\tan \alpha }{ \tan \beta - \tan \alpha}$
Hence Proved

Question:9

If tanθ + secθ =$\l$, then prove that
$sec\theta=\frac{\l ^{2}+1}{2\l }$

Solution. Given : tanθ + secθ =l
There fore
$\frac{\l ^{2}+1}{2\l }=\frac{\left ( \tan \theta +\sec \theta \right )^{2}+1}{2\left ( \tan \theta +\sec \theta \right )}$
$\frac{\tan^{2} \theta +\sec^{2} \theta+2\tan \theta \sec \theta+1}{2\left ( \tan \theta \right )+2\sec \theta }$
$\frac{\sec^{2} \theta +\sec^{2} \theta+2\tan \theta \sec \theta}{2\left ( \tan \theta+\sec \theta \right ) }$ $\left [ \because 1+\tan ^{2}\theta = \sec ^{2}\theta \right ]$
$\frac{2\sec^{2} \theta +2\tan \theta \sec \theta}{2\left ( \tan \theta+2\sec \theta \right ) }$
$\frac{2\sec \theta \left ( \sec \theta+\tan \theta \right ) }{2\left ( \tan \theta+\sec \theta \right ) }$
$= \sec \theta$ (R.H.S)
Hence proved

Question:10

If sinθ + cosθ = p and secθ + cosecθ = q, then prove that q (p2 – 1) = 2p.

Solution. Given :-sinθ + cosθ = p
and secθ + cosecθ = q
To prove :-q (p2 – 1) = 2p
Taking left hand side
q.(p2– 1) =
Put value of q and p we get
$\left ( \sec \theta +\cos ec\theta \right )\left [ \left ( \sin \theta +\cos \theta \right )^{2}-1 \right ]$
$\left ( \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right )$
= $\left ( \frac{1}{\cos \theta +\frac{1}{\sin \theta }} \right )\left [ \left ( \sin^{2} \theta+\cos^{2} \theta+2 \sin \theta \cos \theta \right ) -1\right ]$
$\left ( \because \sec \theta = \frac{1}{\cos \theta } ,\cos ec\theta =\frac{1}{\sin \theta } \right )$
$= \frac{1}{\cos \theta }\left [ \sin ^{2}\theta +\cos ^{2}\theta +2\sin \theta \cos \theta -1 \right ]$
$+ \frac{1}{\sin \theta }\left [ \sin ^{2}\theta +\cos ^{2}\theta +2\sin \theta \cdot \cos \theta -1 \right ]$
$= \frac{\sin ^{2}}{\cos \theta }+\cos \theta+2\sin \theta -\frac{1}{\cos \theta}+\sin \theta+\frac{\cos^{2} \theta}{\sin \theta}+2\cos \theta-\frac{1}{\sin \theta}$
$= 3\cos \theta +3\sin \theta -\frac{1}{\cos \theta}+\frac{\left ( 1+\cos ^{2}\theta \right )}{\cos \theta }+\frac{\left ( 1+\sin ^{2}\theta \right )}{\sin \theta }-\frac{1}{\sin \theta }$
$\left ( \because \sin ^{2}\theta = 1-\cos ^{2}\theta \right )$
$\left ( \because \cos ^{2}\theta = 1-\sin ^{2}\theta \right )$
$= 3\cos \theta +3\sin \theta+\frac{1}{\cos \theta}\times \left ( -1+1-\cos^{2} \theta \right )+\frac{1}{\sin \theta }\times \left ( 1-\sin^{2} \theta-1 \right )$
$\left ( \because \sin^{2} \theta+\cos^{2} \theta= 1 \right )$
$= 3\cos \theta -\cos \theta +3\sin \theta -\sin \theta$
$= 2\cos \theta +2\sin \theta$
$= 2\left ( \cos \theta +\sin \theta \right )$
2p (R.H.S)
Hence proved.

Question:11

If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ =$\sqrt{a^{2}+b^{2}-c^{2}}$

Solution. Given:- asinθ + b cosθ = c
squaring both side we get
$\left ( a\sin \theta +b\cos \theta \right )^{2}= c^{2}$
$a^{2}\sin ^{2}\theta +b^{2}\cos ^{2}\theta+2ab\sin \theta \cos \theta = c^{2}$
$\left ( \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right )$
$\Rightarrow 2ab\sin \theta \cos \theta = c^{2}-a^{2}\sin ^{2}\theta -b^{2}\cos ^{2}\theta \cdots \left ( 1 \right )$
To prove : acosθ – b sinθ =$\sqrt{a^{2}+b^{2}-c^{2}}$
Taking left hand side : a cosθ – b sinθ and square it we get
$\left ( a\cos \theta -b \sin \theta \right )^{2}$
$= a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta -2ab\cos \theta \sin \theta$
$\left [ \because \left ( a-b \right )^{2} = a^{2}+b^{2}-2ab\right ]$
$= a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta -\left ( c^{2}-a^{2}\sin ^{2}\theta -b^{2}\cos ^{2}\theta \right )$ $\text{[Using (1)]}$
$= a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta - c^{2}+a^{2}\sin ^{2}\theta +b^{2}\cos ^{2}\theta$
$= a^{2}\left ( \cos ^{2}\theta+\sin ^{2}\theta \right )+b^{2}\left ( \sin ^{2}\theta+ \cos ^{2}\right )-c^{2}$
$= a^{2}+b^{2} - c^{2}$ $\left ( \because \sin ^{2}\theta +\cos ^{2}\theta = 1 \right )$
Hence $\left ( a\cos \theta -b\sin \theta \right )^{2}= a^{2}+b^{2}-c^{2}$
$\Rightarrow \left ( a\cos \theta -b\sin \theta \right )= \sqrt{a^{2}+b^{2}-c^{2}}$
Hence proved.

Question:12

Prove that
$\frac{1+\sec \theta -\tan \theta }{1+\sec \theta +\tan \theta }= \frac{1-\sin \theta }{\cos \theta }$

Solution To prove :- $\frac{1+\sec \theta -\tan \theta }{1+\sec \theta +\tan \theta }= \frac{1-\sin \theta }{\cos \theta }$

Taking left hand side
$\frac{1+\sec \theta -\tan \theta }{1+\sec \theta +\tan \theta }$
$=\frac{1+\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }}{1+\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }}$ $\begin{bmatrix} \because \because \sec \theta = \frac{1}{\cos \theta } & \\ \tan \theta = \frac{\sin \theta }{\cos \theta }& \end{bmatrix}$
$=\frac{\frac{\cos \theta+1-\sin \theta}{\cos \theta}}{\frac{\cos +1+\sin \theta}{\cos \theta}}$
$\frac{\cos \theta+1-\sin \theta}{\cos \theta+1+\sin \theta}$
Multiply nominator and denominator by (1 – sin $\theta$)
$=\frac{\left ( \cos \theta +1-\sin \theta \right )\left ( 1-\sin \theta \right )}{\left ( \cos \theta +1+\sin \theta \right )\left ( 1-\sin \theta \right )}$
$=\frac{\cos \theta-\cos \theta\sin \theta+1-\sin \theta-\sin \theta+\sin^{2} \theta}{\cos \theta-\sin \theta\cos \theta+1-\sin \theta+\sin \theta-\sin^{2} \theta}$
$=\frac{\cos ec\left ( 1-\sin \theta \right )+\left ( 1-\sin \theta \right )-\sin \left ( 1-\sin \theta \right )}{\cos \theta -\sin \theta \cos \theta+1-\sin ^{2} \theta}$
$=\frac{\left ( 1-\sin \theta \right )\left ( \cos \theta +1-\sin \theta \right )}{\cos \theta-\sin \theta\cos \theta+\cos^{2} \theta}$ $\left ( \because 1-\sin ^{2}\theta = \cos ^{2}\theta \right )$
$=\frac{\left ( 1-\sin \theta \right )\left ( \cos \theta +1-\sin \theta \right )}{\cos \theta\left ( \cos \theta+1-\sin \theta \right )}$
$= \frac{1-\sin \theta}{\cos \theta}$
Hence proved

Question:13

The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the other tower.

Solution. According to the question

Here 30 m is the length of tower AB.
Let h is the height of tower DC
Let the distance between them is x
In $\bigtriangleup$ABC
$\tan 60^{\circ}= \frac{30}{x}$ $\left [ \tan \theta = \frac{Perpendicular}{Base} \right ]$
$\sqrt{3}= \frac{30}{x}$
$x= \frac{30}{\sqrt{3}}\cdots \left ( 1 \right )$
In $\bigtriangleup$BDC
$\tan 30^{\circ}= \frac{h}{x}$
$\frac{1}{\sqrt{3}}= \frac{h}{30}\times \sqrt{3}$ (using (1))
$\frac{30}{\sqrt{3}}=h\sqrt{3}$
$\frac{30}{\sqrt{3}\times\sqrt{3} }= h$
$\frac{30}{3}= h$
h = 10m
Hence the height of the second tower is 10
Distance between them
$=\frac{30}{\sqrt{3}}m$ .

Question:14

From the top of a tower h m high, the angles of depression of two objects, which are in line with the foot of the tower are α and β (β > α). Find the distance between the two objects

According to question

Let x and y are two objects and $\beta$ and $\alpha$ use the angles of depression of two objects.
In $\bigtriangleup$AOX
$\tan \beta = \frac{h}{Ox}$ $\because \tan \theta = \frac{Perpendicular}{Base}$
$Ox= \frac{h}{\tan \beta }$
$Ox= h\cot \beta \cdots \left ( 1 \right )$
In $\bigtriangleup$AOY
$\tan \alpha = \frac{h}{Oy}= \frac{h}{Ox+xy}\; \; \left ( \because Oy= Ox+xy \right )$
$Ox+xy= \frac{h}{\tan \alpha }$
$Ox+xy= h\cot \alpha$
$xy= h\cot \alpha- Ox$
$xy= h\cot \alpha- h\cot \beta$ (from equation (1))
$xy= h\left ( \cot \alpha- \cot \beta \right )$
Hence the distance between two objects is
$h\left ( \cot \alpha- \cot \beta \right )$ .

Question:15

A ladder rests against a vertical wall at an inclination α to the horizontal. Its foot is pulled away from the wall through a distance p so that its upper end slides a distance q down the wall and then the ladder makes an angle β to the horizontal. Show that$\frac{p}{q}= \frac{\cos \beta -\cos \alpha }{\sin \alpha-\sin \beta }$

Solution. According to the question:-

Here a and b be the angles of indication when the ladder at rest and when it pulled away from the wall
In $\bigtriangleup$AOB
$\cos \alpha = \frac{OB}{AB}$ $cos \theta = \frac{Base}{Hypotenuse}$
$OB= AB\cos \alpha \cdots \left ( 1 \right )$
$\sin \alpha = \frac{AO}{AB}$ $\sin \theta = \frac{Perpendicular}{hypotenuse}$

$AO= AB\sin \alpha \cdots \left ( 2 \right )$
Similarly In $\bigtriangleup$DOC
$\cos \beta = \frac{OC}{DC}$
$OC= DC\cos \beta \cdots \left ( 3 \right )$
$\sin \beta = \frac{OD}{DC}$
$OD= DC\sin \beta \cdots \left ( 4 \right )$
Now subtract equation (1) from (3) we get
OC – OB = DC cos$\beta$ – AB cos$\alpha$
Here OC – OB = P
and DC = AB because length of ladder remains $\Rightarrow$ P = AB cos
$\beta$ – AB cos$\alpha$
P = AB (cos$\beta$ – cos$\alpha$) …(5)
Subtract equation (4) from (2) we get
AO – OD = AB sina – DC sin $\beta$
Here AO – OD = q
and AB = DC because length of ladder remains same
$\Rightarrow$ q = AB sin $\alpha$ – AB sin$\beta$
q = AB (sin $\alpha$ – sin $\beta$) …(6)
on dividing equation (5) and (6) we get
$\frac{p}{q}= \frac{AB\left ( \cos \beta -\cos \alpha \right )}{AB\left ( \sin \alpha -\sin \beta \right )}$
$\frac{p}{q}= \frac{\cos \beta -\cos \alpha}{\sin \alpha -\sin \beta}$
Hence proved

Question:16

The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point, 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.

Solution. According to question

Let h is the height of the tower
In $\bigtriangleup$ABE
$\tan \theta = \frac{P}{B}$
$\tan 60^{\circ} = \frac{h}{AB}$ $\left ( \because \theta = 60^{\circ} \right )$
$\sqrt{3}= \frac{h}{AB}$ $\left ( \because \tan 60^{\circ}= \sqrt{3} \right )$
$AB= \frac{h}{\sqrt{3}}\: \cdots \left ( 1 \right )$
In $\bigtriangleup$EDC
$\tan \theta = \frac{P}{B}$
$\tan 45^{\circ} = \frac{h-10}{DC}$
$1= \frac{h-10}{AB}$ $\left ( \because AB-DC,\tan 45^{\circ}= 1 \right )$
$AB= h-10\: \cdots \left ( 2 \right )$
from equation (1) and (2)
$h-10= \frac{h}{\sqrt{3}}$
$\sqrt{3}h-\sqrt{3}10= h$
$\sqrt{3}h-h= \sqrt{3}\times 10$
$h\left ( \sqrt{3}-1 \right )= 10\sqrt{3}$
$h= \frac{10\sqrt{3}}{\sqrt{3}-1}$
Hence the height of the tower
$h= \frac{10\sqrt{3}}{\sqrt{3}-1}m$

Question:17

Solution. According to the question :

Let the height of the other house is X.
In $\bigtriangleup$DCF.
$\tan \alpha = \frac{P}{B}= \frac{EC}{DC}$
$\tan \alpha =\frac{x-h}{DC}$
$DC= \frac{x-h}{\tan \alpha }\cdots \left ( 1 \right )$
In $\bigtriangleup$DAB
. $\tan \beta = \frac{P}{B}= \frac{DA}{AB}$
$\tan \beta = \frac{h}{AB}= \frac{h}{DC}$ $\left ( \because AB= DC \right )$
$\tan \beta = \frac{h}{DC}$
$DC= \frac{h}{\tan \beta }\cdots \left ( 2 \right )$
from equation (1) and (2)
$\frac{X-h}{\tan \alpha}= \frac{h}{\tan \beta}$
$X-h= \frac{\tan \alpha h}{\tan \beta }$
$X= \frac{\tan \alpha h+\tan \beta h}{\tan \beta}$
$X= \frac{h\left ( \tan \alpha +\tan \beta \right )}{\tan \beta}$
separately divide
$X= h\left ( \frac{\tan \alpha }{\tan \beta }+1 \right )$
$X= h\left ( 1+\tan \alpha \cot \beta \right )$ $\left ( \because \frac{1}{\tan \theta }= \cot \theta \right )$
$X= h\left ( 1+\tan \alpha \cot \beta \right )$
Hence proved

Question:18

The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60° and 30°, respectively. Find the height of the balloon above the ground.

Solution

Let y be the height of the balloon from the second window
In $\bigtriangleup$AOB
$\tan 30= \frac{y}{d} \; \; \left ( \because \tan \theta = \frac{Perpendicular}{Base} \right )$
$d= y\sqrt{3}\; \cdots \left ( 1 \right )$ $\left ( \because \tan 30= 1\sqrt{3} \right )$
In $\bigtriangleup$OCD
$\tan 60= \frac{4+y}{d}$
$\sqrt{3}d= 4+y$
$d= \frac{4+y}{\sqrt{3}}\cdots \left ( 2 \right )$
Equating equation (1) & (2) we get
$y\sqrt{3}= \frac{4+y}{\sqrt{3}}$
$y\sqrt{3}\times \sqrt{3}= 4+y$
$3y-y= 4$
$2y= 4$
$y= \frac{4}{2}$
y = 2
Height of balloon = 2 + 4 + y
= 2 + 4 + 2
= 8m

## NCERT Exemplar Solutions Class 10 Maths Chapter 8 Important Topics:

• Trigonometric ratios
• Trigonometric identities
• Proof of some theorems based on these trigonometric identities.
• Trigonometric ratios of complementary angles.
• Class 10 Maths NCERT exemplar chapter 8 solutions discusses the values of trigonometric ratios for some angles as 30° 45° 60° et cetera.
##### Aakash iACST Scholarship Test 2024

Get up to 90% scholarship on NEET, JEE & Foundation courses

## NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

 Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 9 Circles Chapter 10 Constructions Chapter 11 Areas related to Circles Chapter 12 Surface Areas and Volumes Chapter 13 Statistics and Probability

## Features of NCERT Exemplar Class 10 Maths Solutions Chapter 8:

These Class 10 Maths NCERT exemplar chapter 8 solutions provide an introduction to the trigonometric ratios and identities. These trigonometric ratios are defined for the acute angle of the right-angle triangle. Sine X is the ratio of perpendicular side and hypotenuse. Cos X is the ratio of base side and hypotenuse. Tan X is the ratio of perpendicular side and base. A variety of practice problems provided in the exemplar can brush up the skills of the student and enhance the skills related to Trigonometry and its equations based practice problems.

The NCERT exemplar Class 10 Maths Chapter 8 solutions Introduction to Trigonometry and Its Equations consists of a plethora of questions along with detailed solutions and is sufficient to prepare the student to attempt other books such as NCERT Class 10 Maths, RD Sharma Class 10 Maths, RS Aggarwal Class 10 Maths, A textbook of Mathematics by Monica Kapoor et cetera.

NCERT exemplar Class 10 Maths solutions chapter 8 pdf download is a unique feature for the students to provide them with an uninterrupted learning experience as it offers the pdf version of the solutions which can be used while attempting NCERT exemplar Class 10 Maths chapter 8.

### Check Chapter-Wise Solutions of Book Questions

 Chapter No. Chapter Name Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables Chapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry Chapter 9 Some Applications of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related to Circles Chapter 13 Surface Areas and Volumes Chapter 14 Statistics Chapter 15 Probability

### Must, Read NCERT Solution Subject Wise

##### TOEFL ® Registrations 2024

Accepted by more than 11,000 universities in over 150 countries worldwide

##### PTE Exam 2024 Registrations

Register now for PTE & Save 5% on English Proficiency Tests with ApplyShop Gift Cards

### Also, Check NCERT Books and NCERT Syllabus here

1. If we know the value of sine A Then what is the value of cos (90 - A)?

We know that sine is the ratio of perpendicular and hypotenuse.

We know that cos is the ratio of base and hypotenuse.

Therefore, we can say that for complementary angles sine and cosine will give the same values.

2. What is the maximum value of sine?

We know that sine is the ratio of perpendicular and hypotenuse. Hypotenuses cannot have a smaller length on the perpendicular side; hence, the maximum possible value can be one.

3. Is the chapter Introduction to Trigonometry & Its Equations important for Board examinations?

The chapter Introduction to Trigonometry & Its Equations is quite important for Board exams as it carries around 8-10% weightage of the whole paper.

4. How many types of questions from Introduction to Trigonometry & Its Equations appear in the board examination?

Generally, MCQs, Very short, Short, and Long answers type of questions are asked in the board examinations and NCERT exemplar Class 10 Maths solutions chapter 8 are adequate to score well in this chapter.

## Upcoming School Exams

Admit Card Date:01 June,2024 - 20 June,2024

#### National Institute of Open Schooling 12th Examination

Application Date:07 June,2024 - 06 July,2024

#### National Institute of Open Schooling 10th examination

Application Date:07 June,2024 - 06 July,2024

#### Odisha Council of Higher Secondary Education 12th Examination

Others:10 June,2024 - 20 June,2024

#### Nagaland Board of Secondary Education 12th Examination

Exam Date:11 June,2024 - 21 June,2024

Edx
1111 courses
Coursera
792 courses
Udemy
327 courses
Futurelearn
142 courses
IBM
85 courses

## Explore Top Universities Across Globe

University of Essex, Colchester
Wivenhoe Park Colchester CO4 3SQ
University College London, London
Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Bristol, Bristol
Beacon House, Queens Road, Bristol, BS8 1QU
University of Nottingham, Nottingham
University Park, Nottingham NG7 2RD
Lancaster University, Lancaster
Bailrigg, Lancaster LA1 4YW

### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

• Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

• Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

• Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

• Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.

Dear aspirant !

Hope you are doing well ! The class 10 Hindi mp board sample paper can be found on the given link below . Set your time and give your best in the sample paper it will lead to great results ,many students are already doing so .

Hope you get it !

Thanking you

Hello aspirant,

The dates of the CBSE Class 10th and Class 12 exams are February 15–March 13, 2024 and February 15–April 2, 2024, respectively. You may obtain the CBSE exam date sheet 2024 PDF from the official CBSE website, cbse.gov.in.

To get the complete datesheet, you can visit our website by clicking on the link given below.

Thank you

Hope this information helps you.

Hello aspirant,

The paper of class 7th is not issued by respective boards so I can not find it on the board's website. You should definitely try to search for it from the website of your school and can also take advise of your seniors for the same.

You don't need to worry. The class 7th paper will be simple and made by your own school teachers.

Thank you

Hope it helps you.

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9