Continuous practice using ML Aggarwal Class 12 ISC Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.3 can lead to a stronger grasp of mathematical concepts.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 2 Inverse Trigonometric Functions Ex 2.3

Typical problems:

Question 1.
Prove that :
tan-1 $$\sqrt{\frac{x(x+y+z)}{y z}}$$ + tan-1 $$\sqrt{\frac{y(x+y+z)}{z x}}$$ + tan-1 $$\sqrt{\frac{z(x+y+z)}{z x}}$$ = π.
Solution:
L.H.S. = tan-1 $$\sqrt{\frac{x(x+y+z)}{y z}}$$ + tan-1 $$\sqrt{\frac{y(x+y+z)}{z x}}$$ + tan-1 $$\sqrt{\frac{z(x+y+z)}{z x}}$$
put x + y + z = r
= $$\tan ^{-1} \sqrt{\frac{r x}{y z}}+\tan ^{-1} \sqrt{\frac{r y}{x z}}+\tan ^{-1} \sqrt{\frac{r z}{x y}}$$

Question 2.
Solve the following equation for x : 3 tan-1 $$\left(\frac{1}{2+\sqrt{3}}\right)$$ – tan-1 $$\frac{1}{x}$$ = tan-1 $$\frac{1}{2}$$.
Solution:
Given 3 tan-1 $$\left(\frac{1}{2+\sqrt{3}}\right)$$ – tan-1 $$\frac{1}{x}$$ = tan-1 $$\frac{1}{2}$$ …………………(1)
Now 3 tan-1 $$\left(\frac{1}{2+\sqrt{3}}\right)$$ = 3 tan-1 $$\left(\frac{2-\sqrt{3}}{(2+\sqrt{3})(2+\sqrt{3})}\right)$$
= 3 tan-1 $$\left(\frac{2-\sqrt{3}}{4-3}\right)$$
= 3 tan-1 (2 – √3) …………….. (2)
Also tan 15° = tan $$\frac{\pi}{12}$$
= tan (45° – 30°)
= tan $$\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}$$
= $$\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$$
= $$\frac{3+1-2 \sqrt{3}}{3-1}$$
= $$\frac{4-2 \sqrt{3}}{2}$$
= 2 – √3 …………..(3)
From (2) and (3) ; we have
3 tan-1 $$\left(\frac{1}{2+\sqrt{3}}\right)$$ = 3 tan-1 
= 3 × $$\frac{\pi}{12}$$
= $$\frac{\pi}{4}$$ [∵ $$\frac{\pi}{12}$$ ∈ (- $$\frac{\pi}{2}$$, $$\frac{\pi}{2}$$)]
= tan-1 (1)
∴ from (1) : we have
tan-1 (1) – tan-1 ($$\frac{1}{2}$$) = tan-1 $$\frac{1}{x}$$
tan-1 $$\left(\frac{1-\frac{1}{2}}{1+\frac{1}{2} \cdot 1}\right)$$ = tan-1 $$\frac{1}{x}$$
[∵ tan-1x – tan-1 y = tan-1 $$\left(\frac{x-y}{1+x y}\right)$$ if xy > – 1]
⇒ tan-1 $$\left(\frac{\frac{1}{2}}{\frac{3}{2}}\right)$$ = tan-1 $$\frac{1}{x}$$
⇒ tan-1 $$\frac{1}{3}$$ = tan-1 $$\frac{1}{x}$$
⇒ $$\frac{1}{3}=\frac{1}{x}$$
x = 3.

Question 3.
If x = tan-1 (2 tan2 x) – $$\frac{1}{2}$$ sin-1 $$\left(\frac{3 \sin 2 x}{5+4 \cos 2 x}\right)$$, find the greatest values of x.
Solution:
Given, x = tan-1 (2 tan2 x) – $$\frac{1}{2}$$ sin-1 $$\left(\frac{3 \sin 2 x}{5+4 \cos 2 x}\right)$$ …………(*)
⇒ x = tan-1 (2 tan2 x) – $$\frac{1}{2}$$ sin-1 $$\left(\frac{3\left(\frac{2 \tan x}{1+\tan ^2 x}\right)}{5+4\left(\frac{1-\tan ^2 x}{1+\tan ^2 x}\right)}\right)$$
⇒ x = tan-1 (2 tan2 x) – $$\frac{1}{2}$$ sin-1 $$\left(\frac{6 \tan x}{9+\tan ^2 x}\right)$$
⇒ x = tan-1 (2 tan2 x) – $$\frac{1}{2}$$ sin-1 $$\left[\frac{6 \tan x}{9\left(1+\frac{1}{9} \tan ^2 x\right)}\right]$$
⇒ x = tan-1 (2 tan2 x) – $$\frac{1}{2}$$ sin-1 $$\left(\frac{\frac{2}{3} \tan x}{1+\left(\frac{1}{3} \tan x\right)^2}\right)$$ ………………(1)
put $$\frac{1}{3}$$ tan x = tan θ
∴ from (1) ;
⇒ x = tan-1 (2 tan2 x) – $$\frac{1}{2}$$ sin-1 $$\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)$$
⇒ x = tan-1 (2 tan2 x) – $$\frac{1}{2}$$ sin-1 $$\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)$$
⇒ x = tan-1 (2 tan2 x) – $$\frac{1}{2}$$ × 2θ
⇒ x = tan-1 (2 tan2 x) – tan-1 ($$\frac{1}{3}$$ tan x)
⇒ x = tan-1 $$\left[\frac{2 \tan ^2 x-\frac{1}{3} \tan x}{1+\frac{2}{3} \tan ^3 x}\right]$$
⇒ tan x = $$\frac{6 \tan ^2 x-\tan x}{3+2 \tan ^3 x}$$
⇒ tan x [3 + 2 tan3 x – 6 tan x – 1] = 0
⇒ tan x (tan3 x – 3 tan x – 2) = 0
⇒ tan x (tan x – 1)2 (tan x + 2) = 0
∴ tan x = 0 or
(tan x – 1)2 = 0 or
i.e., x = nπ or
x = nπ + $$\frac{\pi}{4}$$ or
tan x = – 2
Since tan x = – 2 does not satisfies the given eqn.
∴ x = nπ, nπ + $$\frac{\pi}{4}$$ ∀ n ∈ I.

Question 4.
Solve the following simultaneous equations:
sin-1 x – sin-1 y = $$\frac{\pi}{3}$$
cos-1 x + cos-1 y = $$\frac{2 \pi}{3}$$
Solution:
Given eqn’s are ;
sin-1 x – sin-1 y = $$\frac{\pi}{3}$$ ………….(1)
cos-1 x + cos-1 y = $$\frac{2 \pi}{3}$$ ……………(2)
Since sin-1 x + cos-1 x = $$\frac{\pi}{2}$$ ∀ x ∈ [- 1, 1]
∴ from (2) ; we have
$$\frac{\pi}{2}$$ – sin-1 x + $$\frac{\pi}{2}$$ – sin-1 y = $$\frac{2 \pi}{3}$$
sin-1 x – sin-1 y = $$\frac{\pi}{3}$$ …………..(3)
On adding (1) and (2) ; we have
2 sin-1 x = $$\frac{2 \pi}{3}$$
⇒ sin-1 x = $$\frac{\pi}{3}$$
⇒ x = sin $$\frac{\pi}{3}$$ = $$\frac{\sqrt{3}}{2}$$
Eqn. (1) – Eqn. (3) ; we have
– 2 sin-1 y = 0
⇒ y = sin 0 = 0
Thus, x = $$\frac{\sqrt{3}}{2}$$ and y = 0.

Question 5.
Find all the positive integral solutions of tan-1 x + cos-1 $$\frac{y}{\sqrt{1+y^2}}$$ = sin-1 $$\frac{3}{\sqrt{10}}$$.
Solution:
Given eqn be,
tan-1 x + cos-1 $$\frac{y}{\sqrt{1+y^2}}$$ = sin-1 $$\frac{3}{\sqrt{10}}$$ …………(1)
First of all, we convert cos-1 to tan-1,
we make right angled ∆ with b = y ;
h = $$\sqrt{1+y^2}$$
∴ p = $$\sqrt{y^2+1-y^2}$$ = 1

Thus, cos-1 $$\frac{y}{\sqrt{1+y^2}}$$ = tan-1 $$\frac{1}{y}$$ …………….(2)
Also, sin-1 $$\left(\frac{3}{\sqrt{10}}\right)$$ = tan-1 $$\frac{3}{1}$$ ………………(3)

Using eqn. (2) and eqn. (3) in eqn. (1) ; we have
tan-1 x + tan-1 y = tan-1 3
⇒ tan-1 $$\frac{1}{y}$$ = tan-1 3 – tan-1 x
= tan-1 $$\left(\frac{3-x}{1+3 x}\right)$$
[∵ tan-1 x – tan-1 y = tan-1 $$\left(\frac{x-y}{1+x y}\right)$$ if xy > – 1]
⇒ $$\frac{1}{y}=\frac{3-x}{1+3 x}$$
⇒ y = $$\frac{1+3 x}{3-x}$$ ……………(4)
But x and y are positive integers
∴ x = 1, 2
when x = 1
∴ from (4) ; y =  = 2
when x = 2
∴ from (4) ; y =  = 7.

Question 6.
If cos-1 x + cos-1 y + cos-1 z = 3π, then find the value of x100 + y100 + z100 – $$\frac{9}{x^{101}+y^{101}+z^{101}}$$.
Solution:
Since 0 ≤ cos-1 x ≤ π,
so maximum value of cos-1 x is π.
∴ cos-1 x + cos-1 y + cos-1 z = 3π
Thus cos-1 x = π ;
cos-1 y = π ;
cos-1 z = π
⇒ x = cos π = – 1 ;
y = cos π = – 1;
z = cos π = – 1
x100 + y100 + z100 – $$\frac{9}{x^{101}+y^{101}+z^{101}}$$ = (- 1)100 + (- 1)100 + (- 1)100 – $$\frac{9}{(-1)^{101}+(-1)^{101}+(-1)^{101}}$$
= 1 + 1 + 1 – $$\frac{9}{-1-1-1}$$
= 1 + 1 + 1 + 3 = 6

Question 7.
If $$\sum_{i=1}^{2 n}$$ sin-1 xi = nπ, then find the value of $$\sum_{i=1}^{2 n}$$ xi.
Solution:
We know that – $$\frac{\pi}{2}$$ ≤ sin-1 x ≤ $$\frac{\pi}{2}$$
Maximum value of sin-1 x be $$\frac{\pi}{2}$$
Since $$\sum_{i=1}^{2 n}$$ sin-1 xi = nπ
⇒ sin-1 xi = $$\frac{\pi}{2}$$ ∀ i = 1, 2, ……………. 2n
⇒ xi = sin $$\frac{\pi}{2}$$ = 1 ∀ i = 1, 2, ……………. 2n
∴ $$\sum_{i=1}^{2 n}$$ xi = (1 + 1 + ………… + 2n terms) = 2n

Question 8.
Find the maximum value of (sin-1 x)3 + (cos-1 x)3.
Solution:
Now, (sin-1 x)3 + (cos-1 x)3
= (sin-1 x + cos-1 x)3 – 3 sin-1 x cos-1 x (sin-1 x + cos-1 x)

Now maximum value of (sin-1 x – $$\frac{\pi}{4}$$) is attained at sin x = – $$\frac{\pi}{2}$$
and Max. value of (sin-1 x – $$\frac{\pi}{4}$$)2 = [- $$\frac{\pi}{2}$$ – $$\frac{\pi}{2}$$]2
= $\frac{3 \pi}{4}$2
∴ from (1) : we have
Thus Max. value of (sin-1 x)3 + (cos-1 x)3 = $$\frac{\pi^3}{32}+\frac{3 \pi}{2} \times \frac{9 \pi^2}{16}$$
= $$\frac{28 \pi^3}{32}=\frac{7 \pi^3}{8}$$.