Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.2 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2
Direction (1 – 4): Prove the following.
Question 1.
3 sin-1 x = sin-1 (3x – 4x3), x ∈ [- \frac{1}{2}, \frac{1}{2}]
Solution.
Let x = sin θ. Then, sin-1 x = θ.
We have,
R.H.S. = sin-1 (3x – 4x3) = sin-1(3 sin θ – 4 sin3 θ)
= sin-1 (sin 3θ) = 3θ = 3 sin-1 x
= L.H.S.
Hence proved.
Question 2.
3 cos-1 x = cos-1 (4x3 – 3x), x ∈ [\frac{1}{2}, 1]
Solution.
Let x = cos θ. Then, cos-1 x = θ.
We have, R.H.S. = cos-1 (4x3 – 3x)
= cos-1 (4cos 3θ – 3 cos θ)
= cos-1 (cos 3θ) = 3θ = 3 cos-1 x
= L.H.S.
Hence proved.
Question 3.
tan-1 \frac{2}{11} + tan-1 \frac{7}{24} = tan-1 \frac{1}{2}.
Solution.
Given, tan-1 \frac{2}{11} + tan-1 \frac{7}{24} = tan-1 \frac{1}{2}
Question 4.
2 tan-1 \frac{1}{2} + tan-1 \frac{1}{7} = tan-1 \frac{31}{17}
Solution.
Given, 2 tan-1 \frac{1}{2} + tan-1 \frac{1}{7} = tan-1 \frac{31}{17}
L.H.S. = 2 tan-1 \frac{1}{2} + tan-1 \frac{1}{7}
= \tan ^{-1}\left[\frac{2 \cdot \frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}\right]+\tan ^{-1}\left(\frac{1}{7}\right) [∵ 2 tan-1 x = tan-1 (\frac{2 x}{1-x^{2}})]
Direction (5 – 10):- Write the following functions in the simplest form:
Question 5.
tan-1 \frac{\sqrt{1+x^{2}}-1}{x}, x ≠ 0.
Solution.
We have, tan-1 \frac{\sqrt{1+x^{2}}-1}{x}
put x = tan θ
⇒ θ = tan-1 x
Question 6.
tan-1 \frac{1}{\sqrt{x^{2}-1}}, |x| > 1
Solution.
Let x = sec θ, then θ = sec-1 x
∴ tan-1 \frac{1}{\sqrt{x^{2}-1}} = tan-1 \left(\frac{1}{\sqrt{\sec ^{2} \theta-1}}\right)
= tan-1 \left(\frac{1}{\sqrt{\tan ^{2} \theta}}\right) [∵ sec2 θ – 1 = tan2 θ]
= tan-1 \left(\frac{1}{\tan \theta}\right)
= tan-1 (cot θ)
= tan-1 [tan (\frac{\pi}{2} – θ)] [∵ tan (\frac{\pi}{2} – θ) = cot θ]
= \frac{\pi}{2} – θ
= \frac{\pi}{2} – sec-1 x
Question 7.
tan-1 \left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right), x < π.
Solution.
We have, tan-1 \left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)
= tan-1 \left(\sqrt{\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}}\right)
= tan-1 (tan \left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right))
= tan-1 (tan \frac{x}{2})
= \frac{x}{2}.
Question 8.
tan-1 (\frac{\cos x-\sin x}{\cos x+\sin x}), 0 < x < π.
Solution.
Question 9.
tan-1 \frac{x}{\sqrt{a^{2}-x^{2}}}, |x| < a.
Solution.
We have, tan-1 \frac{x}{\sqrt{a^{2}-x^{2}}}
Let x = a sin θ
⇒ \frac{x}{a} = sin θ
⇒ θ = sin-1 (\frac{x}{a})
∴ tan-1 \frac{x}{\sqrt{a^{2}-x^{2}}} = tan-1 \left(\frac{a \sin \theta}{\sqrt{a^{2}-a^{2} \sin ^{2} \theta}}\right)
= tan-1 \left(\frac{a \sin \theta}{a \sqrt{1-\sin ^{2} \theta}}\right)
= tan-1 \left(\frac{a \sin \theta}{a \cos \theta}\right)
= tan-1 (tan θ)
= θ = sin-1 \frac{x}{a}.
Question 10.
tan-1 \left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right), a > 0; \frac{-a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}.
Solution.
We have, tan-1 \left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right), a > 0; \frac{-a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}
Let x = a tan θ
⇒ \frac{x}{a} = tan θ
⇒ θ = tan-1 \frac{x}{a}
∴ tan-1 \left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right), a > 0; \frac{-a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}} = tan-1 \left(\frac{3 a^{2} \cdot(a \tan \theta)-a^{3} \tan ^{3} \theta}{a^{3}-3 a \cdot\left(a^{2} \tan ^{2} \theta\right)}\right)
= tan -1 \left(\frac{3 a^{3} \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a^{3} \tan ^{2} \theta}\right)
= tan-1 \left(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right)
= tan-1 (tan 3θ) [∵ tan 3θ = \frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}]
= 3θ = 3 tan-1 \frac{x}{a}.
Direction (11 – 15) : Find the value of each of the following.
Question 11.
tan-1 [2 cos(2 sin-1 \frac{1}{2})].
Solution.
Let sin-1 \frac{1}{2} = x
Then, sin x = \frac{1}{2} = sin (\frac{\pi}{6}))
Now, tan-1 [2 cos(2 sin-1 \frac{1}{2})] = tan-1 [2 cos(2 × \frac{\pi}{6})]
= tan-1 [2 cos \frac{\pi}{3}]
= tan-1 [2 × \frac{1}{2}] [∵ cos (\frac{\pi}{3}) = \frac{1}{2})
= tan-1 1 = \frac{\pi}{4}.
Question 12.
cot(tan-1 a + cot-1 a).
Solution.
We have, cot(tan-1 a + cot-1 a)
= cot (\frac{\pi}{2})
= 0 [∵ tan-1 x + cot-1 x = \frac{\pi}{2}]
Question 13.
tan \frac{1}{2} [sin-1 \frac{2 x}{1+x^{2}} + cos-1 \frac{1-y^{2}}{1+y^{2}}], |x| < 1, y > 0 and xy < 1.
Solution.
Let x = tan θ.
Then, θ = tan-1 x.
∴ sin-1 \frac{2 x}{1+x^{2}} = sin-1 \left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)
= sin-1 (sin 2θ) = 2θ = 2 tan-1 x
Again, let y = tan φ.
Then, φ = tan-1 y
∴ cos-1 \frac{1-y^{2}}{1+y^{2}} = cos-1 \left(\frac{1-\tan ^{2} \varphi}{1+\tan ^{2} \varphi}\right)
= cos-1 (cos 2φ) = 2φ = 2 tan-1 y
Now, tan \frac{1}{2} [sin-1 \frac{2 x}{1+x^{2}} + cos-1 \frac{1-y^{2}}{1+y^{2}}]
= tan \frac{1}{2} [2 tan-1 x + tan-1 y]
= tan [tan-1 x + tan-1 y]
= tan[tan-1 \left(\frac{x+y}{1-x y}\right)]
[∵ tan-1 x + tan-1 y = tan-1 \left(\frac{x+y}{1-x y}\right)]
= \frac{x+y}{1-x y}
Question 14.
If sin(sin-1 \frac{1}{5} + cos-1 x) = 1, then find the value of x.
Solution.
Given, sin(sin-1 \frac{1}{5} + cos-1 x) = 1
⇒ sin-1 \frac{1}{5} + cos-1 x = sin-1 (1)
[∵ sin θ = x ⇒ θ = sin-1 x]
⇒ sin-1 \frac{1}{5} + cos-1 x = sin-1 (sin \frac{\pi}{2})
[∵ sin (\frac{\pi}{2}) = 1]
⇒ sin-1 \frac{1}{5} + cos-1 x = \frac{\pi}{2}
⇒ sin-1 \frac{1}{5} = \frac{\pi}{2} – cos-1 x
sin-1 \frac{1}{5} = sin-1 x
[∵ sin-1 x + cos-1 x = \frac{\pi}{2}]
⇒ \frac{1}{5} = x
Hence, the value of x is \frac{1}{5}.
Question 15.
If tan-1 \frac{x-1}{x-2} + tan-1 \frac{x+1}{x+2}=\frac{\pi}{4}, then find the value of x.
Solution.
We have, tan-1 \frac{x-1}{x-2} + tan-1 \frac{x+1}{x+2}=\frac{\pi}{4}
⇒
Direction (16 – 18): Find the value of each expression.
Question 16.
sin-1 (sin \frac{2 \pi}{3})
Solution.
We have, sin-1 (sin \frac{2 \pi}{3})
We know that sin-1(sin x) = x , if x ∈ (- \frac{\pi}{2}, \frac{\pi}{2}) which is the principal value branch of sin-1 x.
Here, \frac{2 \pi}{3} ∉ (- \frac{\pi}{2}, \frac{\pi}{2}
Now, sin-1 (sin \frac{2 \pi}{3}) can be written as
sin-1 (sin \frac{2 \pi}{3}) = \sin ^{-1}\left[\sin \left(\pi-\frac{\pi}{3}\right)\right]=\sin ^{-1}\left(\sin \frac{\pi}{3}\right) where \frac{\pi}{3} ∈ (- \frac{\pi}{2}, \frac{\pi}{2})
∴ sin-1 (sin \frac{2 \pi}{3}) = sin-1 (sin \frac{\pi}{3}) = \frac{\pi}{3}
Question 17.
tan-1 (tan \frac{3 \pi}{4})
Solution.
We have, tan-1 (tan \frac{3 \pi}{4})
We know that tan-1 (tan x) = x, if x ∈ (- \frac{\pi}{2}, \frac{\pi}{2}), which is the principal value branch of tan-1 x.
Here, \frac{3 \pi}{4} ∉ (- \frac{\pi}{2}, \frac{\pi}{2})
Now, tan-1 (tan \frac{3 \pi}{4}) can be written as
tan-1 (tan \frac{3 \pi}{4}) = tan-1 [tan(π – \frac{\pi}{4})]
= tan-1 [- tan \frac{\pi}{4}]
= tan-1 [tan (- \frac{\pi}{4})]
where – \frac{\pi}{4} ∈ (- \frac{\pi}{2}, \frac{\pi}{2})
[∵ – tan θ = tan(- θ)]
∴ tan-1 (tan \frac{3 \pi}{4}) = tan-1 [tan (- \frac{\pi}{4})]
= – \frac{\pi}{4}.
Question 18.
tan (sin-1 \frac{3}{5} + cot-1 \frac{3}{2})
Solution.
Let sin-1 \frac{3}{5} = x.
Then, sin x = \frac{3}{5}
⇒ cos x = \sqrt{1-\sin ^{2} x} = \frac{4}{5}
⇒ sec x = \frac{5}{4}
∴ tan x = \sqrt{\sec ^{2} x-1}=\sqrt{\frac{25}{16}-1}=\frac{3}{4}
∴ x = tan-1 \frac{3}{4}
∴ sin-1 \frac{3}{5} = tan-1 \frac{3}{4} ………..(i)
Now, cot-1 \frac{3}{2} = tan-1 \frac{2}{3}
[∵ tan-1 \frac{1}{x} = cot-1 x] ……………(ii)
Hence, tan (sin-1 \frac{3}{5} + cot-1 \frac{3}{2})
= tan (tan-1 \frac{3}{4} + tan-1 \frac{2}{3})
= \tan \left(\tan ^{-1} \frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}\right)
[∵ tan-1 x + tan-1 y = tan-1 \left(\frac{x+y}{1-x y}\right)]
= tan (tan-1 \frac{9+8}{12-6})
= tan (tan-1 \frac{17}{6})
= \frac{17}{6}.
Question 19.
cos-1 (cos \frac{7 \pi}{6}) is equal to
(A) \frac{7 \pi}{6}
(B) \frac{5 \pi}{6}
(C) \frac{\pi}{3}
(D) \frac{\pi}{6}
Solution.
We know that cos-1 (cos x) = x if x ∈ [0, x], which is the principal value branch of cos-1 x.
Here, \frac{7 \pi}{6} ∉ x ∈ [0, π]
Now, cos-1 (cos \frac{7 \pi}{6}) can be written as
cos-1 (cos \frac{7 \pi}{6}) = cos-1 [cos(2π – \frac{5 \pi}{6})]
= cos-1 [cos \frac{5 \pi}{6}], where \frac{5 \pi}{6} ∈ [0, π]
[∵ cos(2π – x) = cos x]
∴ cos-1 (cos \frac{7 \pi}{6}) = cos-1 (cos \frac{5 \pi}{6})
= \frac{5 \pi}{6}
The correct option is (B).
Question 20.
sin[\frac{\pi}{3} – sin-1 (- \frac{1}{2})] is equal to
(A) \frac{1}{2}
(B) \frac{1}{3}
(C) \frac{1}{4}
(D) 1
Solution.
Let sin-1 (- \frac{1}{2}) = x.
Then, sin x = – \frac{1}{2} = – sin \frac{\pi}{6} = sin(-\frac{\pi}{6})
We know that the range of the principal value of sin-1 x is (- \frac{\pi}{2}, –\frac{\pi}{2})
∴ sin-1 (- \frac{1}{2}) = – \frac{\pi}{6}
Now, sin[\frac{\pi}{6} – sin-1 (- \frac{1}{2})] = sin \left[\frac{\pi}{3}-\left(-\frac{\pi}{6}\right)\right]
= sin \left(\frac{\pi}{3}+\frac{\pi}{6}\right)
= sin (\frac{3 \pi}{6})
= sin (\frac{\pi}{2}) = 1
The correct option is (D).
Question 21.
tan-1 (- √3) – cot-1 (- √3) is equal to
(A) π
(B) – \frac{\pi}{2}
(C) 0
(D) 2√3
Solution.
Let tan-1 √3 = x
⇒ tan x = √3 = tan \frac{\pi}{3}
∴ tan-1 √3 = \frac{\pi}{3}
Again, let cos-1(- √3) = x
⇒ cot x = – √3 = – cot \frac{\pi}{6}
= cot (π – \frac{\pi}{6})
= cot \frac{5 \pi}{6}
∴ cot-1 (- √3) = \frac{5 \pi}{6}
Now, tan-1 (- √3) – cot-1 (- √3) = \frac{\pi}{3} – \frac{5 \pi}{6}
= \frac{2 \pi-5 \pi}{6}=\frac{-3 \pi}{6}=-\frac{\pi}{2}
Hence, correct option is (B).