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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.

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  1. NCERT Exemplar Solutions Class 10 Maths Chapter 8 Important Topics:
  2. NCERT Class 10 Exemplar Solutions for Other Subjects:
  3. NCERT Class 10 Maths Exemplar Solutions for Other Chapters:
  4. Features of NCERT Exemplar Class 10 Maths Solutions Chapter 8:

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.

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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 \bigtriangleupABC
(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

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. [B]
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}}}

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 \angleC = 90°

In \bigtriangleupABC
\angleA +\angleB +\angleC = 180 [\because sum of interior angles of triangle is 180°]
\angleA + \angleB + 90o = 180 [\because C = 90° (given)]
\angleA + \angleB = 1800 – 900
\angleA + \angleB = 90° …(1)
cos(\angleA +\angleB) = 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 – 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°

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 \bigtriangleupECD
\tan \theta _{1}= \frac{perpendicular}{Base}= \frac{DC}{EC}= \frac{H}{\iota }
Find tan\theta in \bigtriangleupECB 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 \bigtriangleupABC
\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 \bigtriangleupABC
\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 \bigtriangleupABC \angleC = 60°, \angleB = 90°
We know that
\angleA + \angleB + \angleC = 180° (Sum of interior angles of a triangle is 180)
\angleA + 90 + 60 = 180°
\angleA = 30°
In \bigtriangleupABC
\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 \bigtriangleupEDC EC = 20.5 m
DC = 20.5 m
To find angle \theta in \bigtriangleupEDC 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 \bigtriangleupABD
\tan 45^{\circ}= \frac{AB}{BD} \left [ \because \tan \theta = \frac{Perpendicular}{Base} \right ]
1= \frac{AB}{BD}
BD = AB …(1)
In \bigtriangleupABC
\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 \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}

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 \bigtriangleupBDC
\tan \alpha = \frac{\l }{BC} \left [ \because \tan \theta = \frac{Perpendicular}{Base} \right ]
BC= \frac{\l }{\tan \alpha }\cdots \left ( 1 \right )
In \bigtriangleupABC
\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 }

Answer:

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.

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 \bigtriangleupABC
\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 \bigtriangleupBDC
\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 \bigtriangleupAOX
\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 \bigtriangleupAOY
\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 \bigtriangleupAOB
\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 \bigtriangleupDOC
\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 \bigtriangleupABE
\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 \bigtriangleupEDC
\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 \bigtriangleupDCF.
\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 \bigtriangleupDAB
. \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 \bigtriangleupAOB
\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 \bigtriangleupOCD
\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.
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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.

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Frequently Asked Question (FAQs)

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.

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Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

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SWETCHCHA MISKA 10 July,2024
as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

Read Complete Answer
Manasvini Gupta 23 July,2024
i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

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Ashvadha 03 July,2024
i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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sadafreen356 03 July,2024
can we apply for diploma if we have passed 10th cbse lo

Yes, you can definitely apply for diploma courses after passing 10th CBSE. In fact, there are many diploma programs designed specifically for students who have completed their 10th grade.

Generally, passing 10th CBSE with a minimum percentage (often 50%) is the basic eligibility for diploma courses. Some institutes might have specific subject requirements depending on the diploma specialization.

There is a wide range of diploma courses available in various fields like engineering (e.g., mechanical, civil, computer science), computer applications, animation, fashion design, hospitality management, and many more.

You can pursue diplomas at various institutions like:


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Popular CBSE Class 10th Questions

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)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
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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

 
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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)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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
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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

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be a function of the molecular mass of the substance.
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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.
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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
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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
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