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

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 x_i = nπ, then find the value of \(\sum_{i=1}^{2 n}\) x_i.
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 x_i = nπ
⇒ sin^-1 x_i = \(\frac{\pi}{2}\) ∀ i = 1, 2, ……………. 2n
⇒ x_i = sin \(\frac{\pi}{2}\) = 1 ∀ i = 1, 2, ……………. 2n
∴ \(\sum_{i=1}^{2 n}\) x_i = (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
= [latex]\frac{3 \pi}{4}[/latex]²
∴ from (1) : we have
Thus Max. value of (sin^-1 x)³ + (cos^-1 x)³ = \(\frac{\pi^3}{32}+\frac{3 \pi}{2} \times \frac{9 \pi^2}{16}\)
= \(\frac{28 \pi^3}{32}=\frac{7 \pi^3}{8}\).