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

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) $\frac{5}{3}$

Answer:

Answer. [B]
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)
$\left ( AC \right )^{2}= \left ( BC \right )^{2}+\left ( AB \right )^{2}$
$\left ( 5 \right )^{2}= \left ( 4 \right )^{2}+\left ( AB \right )^{2}$
$25-16= \left ( AB \right )^{2}$
$9= \left ( AB \right )^{2}$
$\sqrt{9}= AB$
$3= AB$
The 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

Answer:

$\\sinA=\frac{1}{2}\text{ for angle }30^0\\cosA=\frac{\sqrt{3}}{2}\\cotA=\frac{cosA}{sinA}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$
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)$\frac{3}{2}$

Answer:

Answer. $\quad[B]$
Solution. Given expression is :

$
\begin{aligned}
& {\left[\operatorname{cosec}\left(75^{\circ}+\theta\right)-\sec \left(15^{\circ}-\theta\right)-\tan \left(55^{\circ}+\theta\right)+\cot \left(35^{\circ}-\theta\right)\right]} \\
& {\left[\operatorname{cosec}(90-(15-\theta))-\sec \left(15^{\circ}-\theta\right)-\tan (90-(35-\theta))+\cot \left(35^{\circ}-\theta\right)\right]} \\
& \{\because \text { we can write }(75+\theta)=(90-(15-\theta)) \text { and }(55+\theta)=(90-(35-\theta))\} \\
& \{\because \operatorname{cosec}(90-q)=\sec q \text { and tan }(90-q)=\cot q\} \\
& {[\sec (15-\theta)-\sec (15-\theta)-\cot (35-\theta)+\cot (35-\theta)]=0}
\end{aligned}
$
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)$\frac{a}{\sqrt{b^{2}-a^{2}}}$

Answer:

Answer. [C]
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

Answer:

$\\\text{Given that }\cos(\alpha+\beta)=0\\\Rightarrow (\alpha+\beta)=90^0...........(1)$

$\text{wehave to find }\sin(\alpha-\beta)$

$ \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

Answer:

Answer. [B]
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) $0$

Answer:

Answer. [C]
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
$(A) 0 \ \ \ \ \ (B) 1 \ \ \ \ \ (C)1/2 \ \ \ \ \ (D)\frac{\sqrt{3}}{2}$

Answer:

Answer. [A]
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

Answer:

Answer. [A]
Solution. It is given that sinA + sin²A = 1 …(*)
sinA = 1 – sin²AsinA = 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°

Answer:

Answer. [D]
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

The value of the expression
$\left [ \frac{\sin ^{2}22^{\circ}+\sin ^{2}68^{\circ}}{\cos ^{2}22^{\circ}+\cos ^{2}68^{\circ}}+\sin ^{2}63^{\circ}+\cos ^{2}63^{\circ}\sin 27^{\circ} \right ]\ is$
(A) 3 (B) 2 (C) 1 (D) 0

Answer:

Answer. [B]
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
$(A)\frac{2}{3}$ $(B)\frac{1}{3}$ $(C)\frac{1}{2}$ $(D)\frac{3}{4}$

Answer:
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

Answer:

$\sin\theta-\cos\theta=0$
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

Answer:

Answer. [B]
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

A pole 6 m high casts a shadow 2$\sqrt{3}$ m long on the ground, then the Sun’s elevation is
(A) 60° (B) 45° (C) 30° (D) 90°

Answer:

Answer. [A]
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

Write ‘True’ or ‘False’ and justify your answer in each of the following:
$\frac{\tan 47^{\circ}}{\cot 43^{\circ}}= 1$

Answer:

$\frac{\tan 47^{\circ}}{\cot 43^{\circ}}= 1$
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.

Answer:

Answer. [False]
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.

Answer:

Answer. [False]
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 $\sqrt{\left ( 1-\cos ^{2}\theta \right )\sec ^{2}\theta }= \tan \theta$

Answer:
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.

Answer:

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.

Answer:

(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.

Answer:

Answer. [False]
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.

Answer:

Answer. [False]
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.

Answer:

Answer. [False]
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.

Answer:

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.

Answer:

Answer. [False]
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.

Answer:

Answer. [True]
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
:$\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta}{\sin \theta}= 2\cos ec\theta$

Answer:

$\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta}{\sin \theta}= 2\cos ec\theta$
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 :
$\frac{tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}= 2cosecA$

Answer:

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 .

Answer:

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

Answer:

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

Prove the following :
$\left ( \sqrt{3}+1 \right )\left ( 3-\cot 30^{\circ} \right )= \tan ^{3}60^{\circ}-2\sin 60^{\circ}$
Answer:

$\left ( \sqrt{3}+1 \right )\left ( 3-\cot 30^{\circ} \right )= \tan ^{3}60^{\circ}-2\sin 60^{\circ}$
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 :
$1+\frac{\cot ^{2}\alpha }{1+\cos ec\alpha }= \cos ec\, \alpha$

Answer:

$1+\frac{\cot ^{2}\alpha }{1+\cos ec\alpha }= \cos ec\, \alpha$
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)

Answer:

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.

Answer:

Answer. [30°]
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.

Answer:

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.

Answer:

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

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

Answer:

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

Question:12

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

Answer:

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

Answer:

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.

Answer:

Answer. [45°]
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.

Answer:

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
$cos\theta=\frac{p^{2}-1}{p^{2}+1}$ .

Answer:

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

Answer:

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.

Answer:

Answer. [27.322 m]
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

Answer:

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.

Answer:

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
$\left (2 \sin \theta -\cos \theta \right )^{2}$
$= 4\sin ^{2}\theta +\cos ^{2}\theta -4\sin \theta \cos \theta$

$\left( \because (a - b)^{2} = a^{2} + b^{2} - 2ab \right)$

$= 3\sin ^{2}\theta +\sin ^{2}\theta+\cos ^{2}\theta -4\sin \theta \cos \theta$
$= 3\sin ^{2}\theta +1-4\sin \theta \cos \theta$ $\left [ \because \sin ^{2} \theta +\cos ^{2} \theta= 1\right ]$
$= 3\sin ^{2} \theta+\sin^{2} \theta+4\cos ^{2} \theta$ [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$\sqrt{st}$

Answer:

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.

Answer:

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
$\left ( \frac{h\tan \alpha }{\tan \beta -\tan \alpha } \right )$

Answer:

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 $\triangle B D C$

$
\tan \alpha=\frac{l}{B C}\left[\because \tan \theta=\frac{\text { Perpendicular }}{\text { Base }}\right]
$

$
B C=\frac{l}{\tan \alpha} \cdots
$

In $\triangle \mathrm{ABC}$

$
\tan \beta=\frac{A B}{B C}
$

$
\tan \beta=\frac{h+l}{B C}
$

$
B C=\frac{h+l}{\tan \beta} \cdots
$
Equate equation (1) and (2) we get

$
\begin{aligned}
& \frac{l}{\tan \alpha}=\frac{h+l}{\tan \beta} \\
& l \tan \beta=h \tan \alpha+l \tan \alpha[\text { by cross multiplication] } \\
& l \tan \beta-l \tan \alpha=h \tan \alpha \\
& l(l \tan \beta-\tan \alpha)=h \tan \alpha \\
& l=\frac{h \tan \alpha}{\tan \beta-\tan \alpha}
\end{aligned}
$
Hence Proved

Question:9

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

Answer:

Solution. Given : $\tan \theta+\sec \theta=1$
There fore

$
\begin{aligned}
& \frac{l^2+1}{2 l}=\frac{(\tan \theta+\sec \theta)^2+1}{2(\tan \theta+\sec \theta)} \\
& \frac{\tan ^2 \theta+\sec ^2 \theta+2 \tan \theta \sec \theta+1}{2(\tan \theta)+2 \sec \theta} \\
& \frac{\sec ^2 \theta+\sec ^2 \theta+2 \tan \theta \sec \theta}{2(\tan \theta+\sec \theta)} \quad\left[\because 1+\tan ^2 \theta=\sec ^2 \theta\right] \\
& \frac{2 \sec ^2 \theta+2 \tan \theta \sec \theta}{2(\tan \theta+2 \sec \theta)} \\
& \frac{2 \sec \theta(\sec \theta+\tan \theta)}{2(\tan \theta+\sec \theta)} \\
& =\sec \theta(\text { R.H.S) }
\end{aligned}
$

Hence proved

Question:10

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

Answer:

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 =$\sqrt{a^{2}+b^{2}-c^{2}}$

Answer:

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

Answer:

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.

Answer:

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

Answer:

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$\frac{p}{q}= \frac{\cos \beta -\cos \alpha }{\sin \alpha-\sin \beta }$

Answer:

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.

Answer:

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

A window of a house is h metres above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be α and β, respectively. Prove that the height of the other house is h (1 + tan α cot β) metres.

Answer:

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.

Answer:

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