Effective ML Aggarwal Class 11 Solutions ISC Chapter 3 Trigonometry Chapter Test can help bridge the gap between theory and application.

## ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test

Question 1.
Find the radian measure of an angle (internal) of a regular
(i) pentagon
(ii) hexagon
(iii) polygon of n sides.
Solution:
We know that interior angle of polygon with n sides = $$\left(\frac{2 n-4}{n}\right)$$ × 90°
or $$\left(\frac{2 n-4}{n}\right) \frac{\pi}{2}$$ rad
(i) In pentagon, n = 5
∴ interior angle = $$\left(\frac{2 \times 5-4}{5}\right) \frac{\pi}{2}$$ rad
= $$\frac{6}{5} \times \frac{\pi}{2}$$ radian
= $$\frac{3 \pi}{5}$$ rad.

(ii) In case of hexagon, n = 6
∴ required interior angle = $$\left(\frac{2 \times 6-4}{6}\right) \frac{\pi}{2}$$ rad
= $$\left(\frac{4}{3} \times \frac{\pi}{2}\right)$$ rad
= $$\frac{2 \pi}{3}$$ rad.

(iii) In case of polygon of n sides
∴ required interior angle = $$\left(\frac{2 n-4}{n}\right) \frac{\pi}{2}$$
= $$\left(\frac{n-2}{n}\right)$$ π rad

Question 2.
Find the value of sin (- $$\frac{41 \pi}{4}$$).
Solution:
sin (- $$\frac{4 \pi}{4}$$) = – sin ($$\frac{41 \pi}{4}$$)
[∵ sin (- θ) = – sin θ]
= – sin $$\left(\frac{40 \pi+\pi}{4}\right)$$
= – sin $$\left(10 \pi+\frac{\pi}{4}\right)$$
= – sin $$\frac{\pi}{4}$$
= – 
[.. sin(2nlt+0)sinOVnE 1]

Question 3.
Prove that :
(i) sin2 $$\frac{\pi}{6}$$, sin2 $$\frac{\pi}{4}$$, sin2 $$\frac{\pi}{3}$$ are in A.P.
(ii) tan2 $$\frac{\pi}{6}$$ , tan2 $$\frac{\pi}{4}$$, tan2 $$\frac{\pi}{3}$$ are in GP.
Solution:
(i)

(ii)

Question 4.
Find the value of m2 sin $$\frac{1}{2}$$ π – n2 sin $$\frac{3}{2}$$ π + 2mn sec π.
Solution:
m2 sin $$\frac{1}{2}$$ π – n2 sin $$\frac{3}{2}$$ π + 2mn sec π
= m2 × 1 – n2 sin (π + $$\frac{\pi}{2}$$) + 2mn $$\frac{1}{\cos \pi}$$
= m2 – n2 (9- 1) + 2mn (- 1)
= m2 + n2 – 2mn
= (m – n)2

Question 5.
What is the maximum value of 3 – 7 cos 5x?
Solution:
Since – 1 ≤ cos 5 x ≤ 1
⇒ + 7 ≥ – 7 cos 5x ≥ – 7
⇒ – 7 ≤ – 7 cos 5x ≤ 7
⇒ 3 – 7 ≤ 3 – 7 cos 5x ≤ 3 + 7
⇒ – 4 ≤ 3 – 7 cos 5x ≤ 10
Thus, max. value of 3 – 7 cos 5x = 10

Question 6.
What is the minimum value of 4 + 5 sin (3x – 2) ?
Solution:
Since – 1 ≤ sin (3x – 2) ≤ 1
[∵ |sin θ| ≤ 1 ∀ θ]
⇒ – 5 ≤ 5 sin (3x – 2) ≤ 5
⇒ 4 – 5 ≤ 4 + 5 sin (3x – 2) ≤ 4 + 5
⇒ – 1 ≤ 4 + 5 sin (3x – 2) ≤ 9
Thus, min. value of 4 + 5 sin (3x – 2) = – 1.

Question 7.
What is the maximum value of sin x cos x?
Solution:
Let f(x) = sin x cos x
= $$\frac{1}{2}$$ (2 sin x cos x)
= $$\frac{\sin 2 x}{2}$$
since – 1 ≤ sin 2x ≤ 1
⇒ – $$\frac{1}{2}$$ ≤ $$\frac{1}{2}$$ sin 2x ≤ $$\frac{1}{2}$$
⇒ – $$\frac{1}{2}$$ ≤ f(x) ≤ $$\frac{1}{2}$$
Thus, max. value of f (x) = $$\frac{1}{2}$$.

Question 8.
What is the maximum value of 3 sin x – 4 sin3 x?
Solution:
Let f (x) = 3 sin x – 4 sin3 x
= sin 3x
since – 1 ≤ sin 3x ≤ 1
Thus, max. value of f (x) = 1.

Question 9.
What is the minimum value of 3 cos x – 4 cos3 x?
Solution:
Let f (x) = 3 cos x – 4 cos3 x
= – cos 3x
since, – 1 ≤ cos 3x ≤ 1
⇒ 1 ≥ – cos3x ≥ – 1
⇒ – 1 ≤ — cos 3x ≤ 1
⇒ – 1 ≤ f (x) ≤ 1
Thus, min value of f (x) = – 1.

Question 10.
What is the least value of 2 sin2 x + 3 cos2 x?
Solution:
Let f (x) = 2 sin2 x + 3 cos2 x
= 2 (sin2 x + cos2x) + cos2 x
= 2 + cos2 x
since cos2 x ≥ 0
⇒ 2 + cos2 x ≥ 2
⇒ f (x) ≥ 2
Thus, least value of f (x) = 2.

Question 11.
What is the maximum value of sin x + cos x?
Solution:
Let f (x) = sin x + cos x

Question 12.
What is the minimum value of sin x – cos x?
Solution:
Let f (x) = sin x – cos x
= √2 ($$\frac{1}{\sqrt{2}}$$ sin x – $$\frac{1}{\sqrt{2}}$$ cos x)
= √2 [cos $$\frac{\pi}{4}$$ sin x – sin $$\frac{\pi}{4}$$ cos x]
= √2 sin (x – $$\frac{\pi}{4}$$)
since – 1 ≤ sin (x- $$\frac{\pi}{4}$$) ≤ 1
⇒ – √2 ≤ √2 sin (x – $$\frac{\pi}{4}$$) ≤ √2
⇒ – √2 ≤ f (x) ≤ √2
Thus, min value of f (x) = – √2

Question 13.
If sin 2x = cos 3x and 0 ≤ x < $$\frac{\pi}{2}$$, then find the value of x.
Solution:
Given sin 2x = cos 3x

Question 14.
A railway carriage is travelling along a circular railway track of radius 1500 metres with a speed of 66 km / hour. Find the angle in degrees turned by the engine in 10 seconds.
Solution:
Given radius of circular railway track = r
= 1500 cm
distance covered by a railway carriage in 1 hour = 66 km
So distance covered by a railway carriage in 1 second = $$\frac{66 \times 1000}{60 \times 60}$$
Thus distance covered by a railway carriage in 10 seconds
s = $$\frac{10 \times 60 \times 1000}{60 \times 60}$$ m
= $$\frac{550}{3}$$ m
∴ required angle turned by engine in 10 seconds θ = $$\frac{s}{r}$$
∴ θ = $$\frac{\frac{550}{3}}{1500}$$
= $$\frac{550}{3 \times 1500}=\frac{11}{90}$$ rad
= $$\frac{11}{90} \times \frac{180^{\circ}}{22}$$ × 7
[∵ π rad = 180°]
= 7°

Question 15.
If tan x = $$\frac{a}{b}$$, show that $$\frac{a \sin x-b \cos x}{a \sin x+b \cos x}=\frac{a^2-b^2}{a^2+b^2}$$.
Solution:
Given tan x = $$\frac{a}{b}$$
L.H.S. = $$\frac{a \sin x-b \cos x}{a \sin x+b \cos x}$$
= $$\frac{a \tan x-b}{a \tan x+b}$$
= $$\frac{a\left(\frac{a}{b}\right)-b}{a\left(\frac{a}{b}\right)+b}$$
= $$\frac{a^2-b^2}{a^2+b^2}$$

Question 16.
Is the equation sec2 x – sec x + 1 = 0 possible?
Solution:
Given eqn. be.
6 sec2 x – 5 sec x + 1 = 0
⇒ sec x = 
= 
which is not possible since |sec x| ≥ 1 ∀ x
Hence given eqn. (1) is not possible.

Question 17.
Show that √3 (tan 17° – tan 140°) = 1 + tan 170° tan 140°.
Solution:
Now $$\frac{\tan 170^{\circ}-\tan 140^{\circ}}{1+\tan 170^{\circ} \tan 140^{\circ}}$$ = tan (170° – 140°)
[∵ tan (A – B) = $$\frac{\tan A-\tan B}{1+\tan A \tan B}$$]
= tan 30° = $$\frac{1}{\sqrt{3}}$$
⇒ √3 (tan 17° – tan 140°) = 1 + tan 170° tan 140°

Question 18.
If A + B = 45°, prove that (1 + tan A) (1 + tan B) = 2. Hence, find the value of tan 22 $$\frac{1}{2}^{\circ}$$.
Solution:
Given A + B = 45° ………………(1)
L.H.S. = (1 + tan A) (1 + tan B)
= (1 + tan A) (1 + tan (45° – A)) [using (1)]
= (1 + tan A) $$\left[1+\frac{\tan 45^{\circ}-\tan \mathrm{A}}{1+\tan 45^{\circ} \tan \mathrm{A}}\right]$$
= (1 + tan A) $$\left[1+\frac{1-\tan A}{1+\tan A}\right]$$
= (1 + tan A) $$\left[\frac{1+\tan A+1-\tan A}{1+\tan A}\right]$$
= 2 = R.H.S.

Question 19.
If tan y = $$\frac{Q \sin x}{P+Q \cos x}$$, prove that tan (x – y) = $$\frac{P \sin x}{Q+P \cos x}$$.
Solution:
Given tan y = $$\frac{Q \sin x}{P+Q \cos x}$$ ………………..(1)

Question 20.
Prove that cos2 (x – $$\frac{2 \pi}{3}$$) + cos2 (x + $$\frac{2 \pi}{3}$$) = $$\frac{3}{2}$$
Solution:

Question 21.
If x, y and z are in A.P., prove that y = $$\frac{\sin x-\sin z}{\cos z-\cos x}$$.
Solution:
Since x, y, z are in A.P.
∴ y – x = z – y
⇒ 2y = x + z

Question 22.
If sin 2x = λ sin 2y, prove that $$\frac{\tan (x+y)}{\tan (x-y)}=\frac{\lambda+1}{\lambda-1}$$.
Solution:
Given sin 2x = λ sin 2y

Question 23.
Prove that $$\frac{1-\sin 2 x}{1+\sin 2 x}=\tan ^2\left(\frac{\pi}{4}-x\right)$$.
Solution:

Question 24.
Prove that $$\sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}=\frac{3}{2}$$.
Solution:

Question 25.
Given that sin x = – $$\frac{3}{5}$$, cos y = – $$\frac{5}{13}$$ and x is in the same quadrant asy, evaluate without using tables :
(i) cos (x – y)
(ii) tan (x + y)
(iii) cos $$\frac{x}{2}$$
Solution:
Given sin x = – $$\frac{3}{5}$$
and cos y = – $$\frac{5}{13}$$
Since sin x and cos y both are negative.
∴x and y lies in 3rd quadrant as x and y lies in same quadrant.
So $$\frac{x}{2}$$, $$\frac{y}{2}$$ lies in second qu1drant.
∴ cos x = – $$\sqrt{1-\sin ^2 x}$$
= $$-\sqrt{1-\left(-\frac{3}{5}\right)^2}$$
= $$-\sqrt{1-\frac{9}{25}}$$
= $$-\sqrt{\frac{16}{25}}=-\frac{4}{5}$$
and sin y = – $$\sqrt{1-\cos ^2 y}$$
= $$-\sqrt{1-\left(\frac{5}{13}\right)^2}$$
= $$-\sqrt{1-\frac{25}{169}}$$
= $$-\sqrt{\frac{144}{169}}=-\frac{12}{13}$$

(i) cos (x – y) = cos x cos y + sin x sin y
= $$\left(-\frac{4}{5}\right)\left(-\frac{5}{13}\right)+\left(-\frac{3}{5}\right)\left(-\frac{12}{13}\right)$$
= $$\frac{20+36}{65}=\frac{56}{65}$$

(ii) tan (x + y) = $$\frac{\tan x+\tan y}{1-\tan x \tan y}$$
= $$\frac{\frac{3}{4}+\frac{12}{5}}{1-\frac{3}{4} \times \frac{12}{5}}$$
= $$\frac{\frac{15+48}{20}}{1-\frac{36}{20}}$$
= $$\frac{63}{-16}$$
[tan x = $$=\frac{\sin x}{\cos x}$$
= $$\frac{-\frac{3}{5}}{-\frac{4}{5}}=\frac{3}{4}$$
and tan y = $$\frac{\sin y}{\cos y}$$
= $$\frac{-\frac{12}{13}}{-\frac{5}{13}}=\frac{12}{5}$$]

(iii) since $$\frac{x}{2}$$ lies in second quadrant
∴ cos x < 0
Thus, cos $$\frac{x}{2}$$ = – $$\sqrt{\frac{1+\cos x}{2}}$$
= – $$\sqrt{\frac{1+\left(-\frac{4}{5}\right)}{2}}$$
= $$-\sqrt{\frac{1}{10}}=-\frac{1}{\sqrt{10}}$$

Question 26.
Prove that :
(i) $$\frac{1-\cos x+\cos y-\cos (x+y)}{1+\cos x-\cos y-\cos (x+y)}=\tan \frac{x}{2} \cot \frac{y}{2}$$
(ii) $$\frac{\cos ^3 x-\cos 3 x}{\cos x}+\frac{\sin ^3 x+\sin 3 x}{\sin x}$$ = 3.
Solution:
(i)

(ii)

Question 27.
Prove that $$\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}$$ = 4.
Solution:

Question 28.
Show that never lies between $$\frac{1}{3}$$ and 3.
Solution:
Let y = $$\frac{\tan 3 x}{\tan x}$$
= $$\frac{3 \tan x-\tan ^3 x}{\tan x\left(1-3 \tan ^2 x\right)}$$
= $$\frac{3-\tan ^2 x}{1-3 \tan ^2 x}$$
⇒ y (1 – 3 tan2 x) = 3 – tan2 x
⇒ tan2 x (1 – 3y) = 3 – y
⇒ tan2 x = $$\frac{3-y}{1-3 y}$$
But tan2 x ≥ 0 ∀ x ∈ R
⇒ $$\frac{3-y}{1-3 y}$$ ≥ 0
⇒ (3 – y) (1 – 3y) ≥ 0
[∵ (1 – 3y)2 ≥ 0 ∀ x ∈ R]
⇒ (y – 3) (3y – 1) ≥ 0
⇒ (y – 3) (y – $$\frac{1}{3}$$) ≥ 0
⇒ y ≥ 3 or y ≤ $$\frac{1}{3}$$
[∵ (x – a) (x – b) ≥ 0
and a > b = x ≥ a or x ≤ b]
Thus y i.e. $$\frac{\tan 3 x}{\tan x}$$ never lies between $$\frac{1}{3}$$ and 3.

Question 29.
Solve the following equations :
(i) tan 2x = – cot (x + $$\frac{\pi}{6}$$)
(ii) cot2 x + 3 cosec x + 3 = 0
(iii) 4 sin2 x + √3 = 2 (1 + √3) sin x
(iv) tan2 x – (1 + √3) tan x + √3 = 0
(v) cos 2x – cos 8x + cos 6x = 1
(vi) tan ($$\frac{\pi}{4}$$ + x) + tan ($$\frac{\pi}{4}$$ – x) = 4
(vii) cosec x = 1 + cot x
Solution:
(i) tan 2x = – cot (x + $$\frac{\pi}{6}$$)
⇒ tan 2x = tan $$\left(\frac{\pi}{2}+x+\frac{\pi}{6}\right)$$
= tan $$\left(\frac{2 \pi}{3}+x\right)$$
⇒ 2x = nπ + $$\frac{2 \pi}{3}$$ + x ; n ∈ I
[∵ tan θ = tan α
⇒ θ = nπ + α, n ∈ I]
⇒ x = nπ + $$\frac{2 \pi}{3}$$ ; n ∈ I

(ii) Given eqn. be,
cot2 x + 3 cosec x + 3 = 0
⇒ cosec2 x – 1 + 3 cosec x + 3 = 0
⇒ cosec2 x + 3 cosec x + 2 = 0
⇒ cosec x = $$\frac{-3 \pm \sqrt{9-8}}{2}$$
= $$\frac{-3 \pm 1}{2}$$
= – 2, – 1
either cosec x = – 2 or cosec x = – 1
sin x = – $$\frac{1}{2}$$
or sin x = – 1
⇒ sin x = – sin $$\frac{\pi}{6}$$
or sin x = sin (- $$\frac{\pi}{2}$$)
⇒ x = nπ + (- 1)n (- $$\frac{\pi}{6}$$)
or x = nπ + (- 1)n (- $$\frac{\pi}{2}$$)
Hence required solutions of given eqn. are
x = nπ – (- 1)n (- $$\frac{\pi}{6}$$) or nπ – (- 1)n (- $$\frac{\pi}{2}$$)

(iii) Given eqn. be,
4 sin2 x + √3 = 2 (1 + √3) sin x

(iv) Given eqn. be,
tan2 x – (1 + √3) tan x + √3 = 0
which is quadratic in tan x.
∴ tan x = $$\frac{(1+\sqrt{3}) \pm \sqrt{(1+\sqrt{3})^2-4 \sqrt{3}}}{2}$$
= $$\frac{(1+\sqrt{3}) \pm \sqrt{(1-\sqrt{3})^2}}{2}$$
= $$\frac{(1+\sqrt{3}) \pm(1-\sqrt{3})}{2}$$
= 1, √3
either tan x = 1 or tan x = √3
tan x = 1 = tan $$\frac{\pi}{4}$$
tan x = √3 = tan $$\frac{\pi}{3}$$
⇒ x = nπ + $$\frac{\pi}{4}$$
or x = nπ + $$\frac{\pi}{3}$$ ∀ n ∈ I
Hence required solutions are nπ + $$\frac{\pi}{4}$$, nπ + $$\frac{\pi}{3}$$ ; n ∈ I

(v) Given eqn.be
cos 2x – cos 8x + cos 6x = 1
⇒cos 2x – cos 8x – (1 – cos 6x) = 0
⇒ 2 sin $$\left(\frac{2 x+8 x}{2}\right)$$ sin $$\left(\frac{8 x-2 x}{2}\right)$$ – 2 sin2 3x = 0
⇒ 2 sin 5x sin 3x – 2 sin2 3x = 0
⇒ 2 sin 3x [sin 5x – sin 3x] = 0
⇒ 2 sin 3x (2 cos 4x sin x) = 0
either sin 3x = 0 or cos 4x =0 or sin x = 0
3x = nπ or 4x = (2n + 1) $$\frac{\pi}{2}$$
or x = nπ ; n ∈ I
⇒ x = $$\frac{n \pi}{3}$$ or x = (2n + 1) $$\frac{\pi}{8}$$
or x = nπ ; n ∈ I
Since the solution x = nπ
i.e. x = π, 2π, 3π ……………….. are included in the
solution x = $$\frac{n \pi}{3}$$ for n = 3, 6, 9 ……………..
Hence, the required solutions are
x = nπ, (2n + 1) $$\frac{\pi}{8}$$ ∀ n ∈ I

(vi) Given tan $$\left(\frac{\pi}{4}+x\right)$$ + tan $$\left(\frac{\pi}{4}-x\right)$$ = 4
⇒ $$\frac{\tan \frac{\pi}{4}+\tan x}{1-\tan \frac{\pi}{4} \tan x}+\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \tan x}$$ = 4
⇒ $$\frac{1+\tan x}{1-\tan x}+\frac{1-\tan x}{1+\tan x}$$ = 4
⇒ (1 + tan x)2 + (1 – tan x)2 = 4 (1 – tan2 x)
⇒ 2 tan2 x + 2 = 4 (1 – tan2 x)
⇒ 6 tan2 x = 2
⇒ tan2 x = $$\frac{1}{3}=\left(\frac{1}{\sqrt{3}}\right)^2$$
= tan2 $$\frac{\pi}{6}$$
⇒ x = nπ ± $$\frac{\pi}{6}$$ ∀ n ∈ I
[∵ tan2 θ = tan2 α
⇒ nπ ± α ∀ n ∈ I]

(vii) Given cosec x = 1 + cot x
⇒ $$\frac{1}{\sin x}=1+\frac{\cos x}{\sin x}$$
⇒ 1 = sin x + cos x ; sin x ≠ 0 …………….(1)
which is of the form a cos x + b sin x = c
dividing throughout eqn. (1) by $$\sqrt{1^2+1^2}$$ i.e. √2 ; we get
$$\frac{1}{\sqrt{2}}$$ cos x + $$\frac{1}{\sqrt{2}}$$ sin x = $$\frac{1}{\sqrt{2}}$$

⇒ x = 2nπ + $$\frac{\pi}{2}$$, 2nπ ; n ∈ I
As sin (2n∈) = 0 ∀ n ∈ I but sin x ≠ 0
Hence the required solutions are 2nπ + $$\frac{\pi}{2}$$ ; n ∈ I

Question 30.
In any triangle ABC, prove that
(i) 2 (b cos2 $$\frac{C}{2}$$ + c cos2 $$\frac{B}{2}$$) = a + b + c
(ii) $$\frac{\sin \mathrm{A}}{\sin (A+B)}=\frac{a}{c}$$
(iii) $$\frac{a-b}{a+b}=\frac{\tan \frac{A-B}{2}}{\tan \frac{A+B}{2}}$$
(iv) $$\frac{b+c}{b-c}=\cot \frac{A}{2} \cot \frac{B-C}{2}$$
(v) $$\frac{1+\cos (\mathrm{A}-\mathrm{B}) \cos \mathrm{C}}{1+\cos (\mathrm{A}-\mathrm{C}) \cos \mathrm{B}}=\frac{a^2+b^2}{a^2+c^2}$$
Solution:
(i) L.H.S. = 2 $$\left(b \cos ^2 \frac{\mathrm{C}}{2}+c \cos ^2 \frac{\mathrm{B}}{2}\right)$$
= 2 $$\left[b\left(\frac{1+\cos \mathrm{C}}{2}\right)+c\left(\frac{1+\cos \mathrm{B}}{2}\right)\right]$$
= [b + c + b cos C + c cos B]
= b + c + a [using projection formulae]
= R.H.S.

(ii) L.H.S. = $$\frac{\sin A}{\sin (A+B)}$$
= $$\frac{\sin A}{\sin (\pi-C)}$$
[∵ A + B + C = π
⇒ A + B = π – C]
= $$\frac{\sin \mathrm{A}}{\sin \mathrm{C}}=\frac{a}{c}$$
[by sine formula, $$\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}$$
⇒ $$\frac{\sin \mathrm{A}}{\sin \mathrm{C}}=\frac{a}{c}$$]
= R.H.S.

(iii)

(iv)

(v)

Question 31.
In a triangle ABC, if cos A + cos C = sin B, then prove that it is right anglcd triangle.
Solution:
Given cos A + cos C = sin B

either sin $$\frac{B}{2}$$ = 0 or cos C = 0 or cos A = 0
i.e. $$\frac{B}{2}$$ = 0 or C = 90° or A = 90°
i.e. B = 0 or C = 90° or A = 90°
either C = 90° or A = 90° [∵ B ≠ 0]
Hence ∆ is right angled triangle.