Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise Textbook Exercise Questions and Answers.
PSEB Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise
Find the value of the following :
Question 1.
cos-1 (cos \frac{13 \pi}{6}).
Solution.
We know that cos-1(cos x) = x if x ∈ [0, π], which is the principal value of cos-1 x.
Here, \frac{13 \pi}{6} ∉ [0, π].
Now, cos-1 (cos \frac{13 \pi}{6}) can be written as
cos-1 (cos \frac{13 \pi}{6}) = cos-1 [cos (2π + \frac{pi}{6})]
= cos-1 [cos (\frac{pi}{6})],
where \frac{pi}{6} ∈ [0, π]
[∵ cos(2π + x) = cos x]
∴ cos-1 (cos \frac{13 \pi}{6}) = cos-1 [cos (\frac{pi}{6})]
= \frac{pi}{6}.
Question 2.
tan-1 (tan \frac{7 \pi}{6})
Solution.
We know that tan-1(tan x) = x if x ∈ (-\frac{\pi}{2}, \frac{\pi}{2}) which is the principal value of cos-1 x.
Here, \frac{7 \pi}{6} ∉ (-\frac{\pi}{2}, \frac{\pi}{2})
Now, tan-1 (tan \frac{7 \pi}{6}) can be written as
tan-1 (tan \frac{7 \pi}{6}) = tan-1 (tan (π + \frac{\pi}{6}))
= tan-1 [tan (\frac{\pi}{6})]
where \frac{pi}{6} ∈ (-\frac{\pi}{2}, \frac{\pi}{2})
[∵ tan(π + x) = tan x]
∴ tan-1 (tan \frac{7 \pi}{6}) = tan-1 [tan (\frac{\pi}{6})]
= \frac{pi}{6}
Prove that
Question 3.
2 sin-1 \frac{3}{5} = tan-1 \frac{24}{7}.
Solution.
Let sin-1 \frac{3}{5} = x.
Then, sin x = \frac{3}{5}
⇒ cos x = \sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{25-9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}
∴ tan x = \frac{3 / 5}{4 / 5}=\frac{3}{4}
∴ x = tan-1 \frac{3}{4}
⇒ sin-1 \frac{3}{5} = tan-1 \frac{3}{4}
Now, we have
L.H.S = 2 sin-1 \frac{3}{5} = 2 tan-1 \frac{3}{4}
= tan-1 \left(\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right)
[∵ 2 tan-1 x = tan-1 \frac{2 x}{1-x^{2}}]
= tan-1 \left(\frac{\frac{3}{2}}{\frac{16-9}{16}}\right)
= tan-1 \left(\frac{3}{2} \times \frac{16}{7}\right)
= tan-1 \frac{24}{7}
= R.H.S.
Hence proved.
Question 4.
sin-1 \frac{8}{17} + sin-1 \frac{3}{5} = tan-1 \frac{77}{36}
solution.
Let sin-1 \frac{8}{17} = x.
Then, sin x = \frac{8}{17}
⇒ cos x = \sqrt{1-\left(\frac{8}{17}\right)^{2}}=\sqrt{\frac{225}{289}}=\frac{15}{17}
∴ tan x = \frac{8 / 17}{15 / 17}=\frac{8}{15}
⇒ x = tan-1 \frac{8}{15}
∴ sin-1 \frac{8}{17} = tan-1 \frac{8}{15} …………..(i)
Now, let sin-1 \frac{3}{5} = y.
Then, sin y = \frac{3}{5}
⇒ cos y = \sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}
∴ tan y = \frac{3 / 5}{4 / 5}=\frac{3}{4}
⇒ y = tan-1 \frac{3}{4}
∴ sin-1 \frac{3}{5} = tan-1 \frac{3}{4} …………..(ii)
Now, we have
L.H.S = sin-1 \frac{8}{17} + sin-1 \frac{3}{5}
[Using Eqs. (i) and (ii)]
= tan-1 \frac{8}{15} + tan-1 \frac{3}{4}
= tan-1 \frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15} \times \frac{3}{4}}
= tan-1 \left(\frac{32+45}{60-24}\right)
[tan-1 x + tan-1 y = tan-1 \left(\frac{x+y}{1-x y}\right)]
= tan-1 \frac{77}{36}
= R.H.S
Hence proved.
Question 5.
cos-1 \frac{4}{5} + cos-1 \frac{12}{13} = cos-1 \frac{33}{65}
Solution.
Let cos-1 \frac{4}{5} = x
Then, cos x = \frac{4}{5}
⇒ sin x = \sqrt{1-\left(\frac{4}{5}\right)^{2}}=\sqrt{\frac{9}{25}}=\frac{3}{5}
∴ tan x = \frac{3 / 5}{4 / 5}=\frac{3}{4} …………(i)
⇒ x = tan-1 \frac{3}{4}
Now, let cos-1 \frac{12}{13} = y.
Then cos y = \frac{12}{13}
⇒ sin y = \sqrt{1-\left(\frac{12}{13}\right)^{2}}=\sqrt{\frac{25}{169}}=\frac{5}{13}
∴ tan y = \frac{5 / 13}{12 / 13}=\frac{5}{12}
⇒ y = tan-1 \frac{5}{12}
∴ cos-1 \frac{12}{13} = tan-1 \frac{5}{12} ……………(ii)
Let cos-1 \frac{33}{65} = z.
Then, cos z = \frac{33}{65}
⇒ sin z = \sqrt{1-\left(\frac{33}{65}\right)^{2}}=\sqrt{\frac{3136}{4225}}=\frac{56}{65}
∴ tan z = \frac{56 / 65}{33 / 65}=\frac{56}{33}
⇒ z = tan-1 \frac{56}{33}
∴ cos-1 \frac{33}{65} = tan-1 \frac{56}{33} …………..(iii)
Now, we have
L.H.S = cos-1 \frac{4}{5} + cos-1 \frac{12}{13}
= \frac{3 / 5}{4 / 5}=\frac{3}{4} + tan-1 \frac{5}{12}
[∵ Usin Eqs. (i) and (ii)]
= tan-1 \frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \cdot \frac{5}{12}}
[∵ tan-1 x + tan-1 y = tan-1 \left(\frac{x+y}{1-x y}\right)]
= tan-1 \frac{36+20}{48-15}
= tan-1 \frac{56}{33}
= cos-1 \frac{33}{65} = R.H.S.
Hence proved.
Question 6.
cos-1 \frac{12}{13} + sin-1 \frac{3}{5} = sin-1 \frac{56}{65}
Solution.
Let sin-1 \frac{3}{5} = x.
Then, sin x = \frac{3}{5}
⇒ cos x= \sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}
∴ tan x = \frac{3 / 5}{4 / 5}=\frac{3}{4}
⇒ x = tan-1 \frac{3}{4}
∴ sin-1 \frac{3}{5} = tan-1 \frac{3}{4} …………(i)
Now, let cos-1 \frac{12}{13} = y.
Then, cos y = \frac{12}{13}
⇒ sin y = \sqrt{1-\left(\frac{12}{13}\right)^{2}}=\sqrt{\frac{25}{169}}=\frac{5}{13}
∴ tan y = \frac{5 / 13}{12 / 13}=\frac{5}{12}
⇒ y = tan-1 \frac{5}{12}
∴ cos-1 \frac{12}{13} = tan-1 \frac{5}{12} ………………(ii)
Let sin-1 \frac{56}{65} = z.
Then, sin z = \frac{56}{65}
⇒ cos z = \sqrt{1-\left(\frac{56}{65}\right)^{2}}=\sqrt{\frac{1089}{4225}}=\frac{33}{65}
∴ tan z = \frac{56 / 65}{33 / 65}=\frac{56}{33}
⇒ z = tan-1 \frac{56}{33}
∴ sin-1 \frac{56}{65} = tan-1 \frac{56}{33} …………….(iii)
Now, we have
L.H.S. = cos-1 \frac{12}{13} + sin-1 \frac{3}{5}
= tan-1 \frac{3}{4} + tan-1 \frac{5}{12}
[Using Eqs. (i) and (ii)]
= tan-1 \frac{\frac{5}{12}+\frac{3}{4}}{1-\frac{5}{12} \cdot \frac{3}{4}}
[∵ tan-1 x + tan-1 y = tan-1 \left(\frac{x+y}{1-x y}\right)]
= tan-1 \frac{20+36}{48-15}
= tan-1 \frac{56}{33}
= sin-1 \frac{56}{65}.
Hence proved.
Question 7.
tan-1 \frac{63}{16} = sin-1 \frac{5}{13} + cos -1 \frac{3}{5}
Solution.
Let sin-1 \frac{5}{13} = x
Then, sin x = \frac{5}{13}
⇒ cos x = \sqrt{1-\left(\frac{5}{13}\right)^{2}}=\sqrt{\frac{144}{169}}=\frac{12}{13}
∴ tan x = \frac{5 / 13}{12 / 13}=\frac{5}{12}
⇒ x = tan-1 \frac{5}{12}
∴ sin-1 \frac{5}{13} = tan-1 \frac{5}{12} ………….(i)
Let cos-1 \frac{3}{5} = y.
Then, cos y = \frac{3}{5}
⇒ sin y = \sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}
∴ tan y = \sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}
⇒ y = tan-1 \frac{4}{5}
∴ cos-1 \frac{3}{5} = tan-1 \frac{4}{5} ………………(ii)
Using Eqs. (i) and (ii), we have
R.H.S = sin-1 \frac{5}{13} + cos-1 \frac{3}{5}
= tan-1 \frac{5}{12} + tan-1 \frac{4}{5}
[∵ tan-1 x + tan-1 y = tan-1 \left(\frac{x+y}{1-x y}\right)]
= tan-1 \left(\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \times \frac{4}{3}}\right)
= tan-1 \left(\frac{15+48}{36-20}\right)
= tan-1 \frac{63}{16} = L.H.S.
Hence proved.
Question 8.
tan-1 \frac{1}{5} + tan-1 \frac{1}{7} + tan-1 \frac{1}{3} + tan-1 \frac{1}{8} = \frac{\pi}{4}
Solution.
tan-1 \frac{1}{5} + tan-1 \frac{1}{7} + tan-1 \frac{1}{3} + tan-1 \frac{1}{8}
Question 9.
tan-1 √x = \frac{1}{2} cos-1 \left(\frac{1-x}{1+x}\right), x ∈ [0, 1]
Solution.
Let x = tan2 θ.
Then, √x = tan θ
⇒ θ = tan-1 √x
∴ \frac{1-x}{1+x}=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} = cos 2θ
[∵ cos 2θ = \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}]
Now, we have
R.H.S = \frac{1}{2} cos-1 \left(\frac{1-x}{1+x}\right)
= \frac{1}{2} cos-1 2θ = θ
= tan-1 √x = L.H.S
Hence Proved.
Question 10.
cot-1 \left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}, x ∈ (0, \frac{\pi}{4})
Solution.
Consider, \left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)
= \frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^{2}}{(\sqrt{1+\sin x})^{2}-(\sqrt{1-\sin x})^{2}} (By rationalising)
= \frac{(1+\sin x)+(1-\sin x)+2 \sqrt{(1+\sin x)(1-\sin x)}}{1+\sin x-1+\sin x}
= \frac{2\left(1+\sqrt{\left.1-\sin ^{2} x\right)}\right.}{2 \sin x}=\frac{1+\cos x}{\sin x}=\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}
= cot \frac{x}{2}
L.H.S = cot-1 \left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)
= cot-1 (cot \frac{x}{2})
= \frac{x}{2} = R.H.S
Question 11.
tan-1 \left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right) = \frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x, -\frac{1}{\sqrt{2}} ≤ x ≤ 1.
[Hint: put x = cos 2θ]
Solution.
Put x = cos 2θ, so that θ = \frac{1}{2} cos-1 x.
Then, we have
Question 12.
\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}
Solution.
L.H.S = \frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}
= \frac{9}{4}\left(\frac{\pi}{2}-\sin ^{-1} \frac{1}{3}\right)
= =\frac{9}{4}\left(\cos ^{-1} \frac{1}{3}\right) [∵ sin-1 x + cos-1 x = \frac{\pi}{2}]
= \frac{9}{4}\left(\sin ^{-1} \sqrt{1-\left(\frac{1}{3}\right)^{2}}\right)
[∵ cos-1 x = sin-1 \sqrt{1-x^{2}}]
= \frac{9}{4} \sin ^{-1} \sqrt{\frac{8}{9}}
= \frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}
= R.H.S
Hence proved.
Direction (13 – 17): Solve the following equations.
Question 13.
2 tan-1 (cos x) = tan-1 (2cosec x)
Solution.
We have, 2 tan-1 (cos x) = tan-1 (2 cosec x)
⇒ tan-1 \left(\frac{2 \cos x}{1-\cos ^{2} x}\right) = tan-1 (2 cosec x)
[∵ 2 tan-1 x = tan-1 \frac{(2 x)}{1-x^{2}}]
⇒ \left(\frac{2 \cos x}{1-\cos ^{2} x}\right) = 2 cosec x
⇒ \frac{2 \cos x}{\sin ^{2} x}=\frac{2}{\sin x}
⇒cos x = sin x
⇒ tan x = 1 = tan \frac{\pi}{4}.
Question 14.
tan-1 \frac{1-x}{1+x} = \frac{1}{2} tan-1 x, (x > 0)
Solution.
We have, sin-1 (1 – x) – 2 sin-1 x = \frac{\pi}{2}
⇒ – 2 sin-1 x = \frac{\pi}{2} – sin-1 (1 – x)
⇒ – 2 sin-1 x = cos-1 (1 – x) ………….(i)
Let sin-1 x = θ
⇒ sin θ = x
⇒ cos θ = \sqrt{1-x^{2}}
∴ θ = cos-1 \sqrt{1-x^{2}}
∴ sin-1 x = cos-1 \sqrt{1-x^{2}}
Therefore, from Eq. (i), we have
– 2 cos-1 (\sqrt{1-x^{2}} ) = cos-1 (1 – x)
Put x = sin y. Then, we have
– 2 cos-1 () = cos-1 (1 – sin y)
⇒ – 2 cos-1 (cos y) = cos-1 (1 – sin y)
⇒ – 2y = cos-1 (1 – sin y)
⇒ 1 – sin y = cos(- 2y) = cos 2y
⇒ 1 – sin y = 1 – 2 sin2 y
⇒ 2 sin2 y – sin y = 0
⇒ sin y(2 sin y – 1) = 0
sin y = 0 or \frac{1}{2}
∴ x = 0 or x = \frac{1}{2}
But, when x = \frac{1}{2}, it can be observed that
We have, tan-1 \frac{1-x}{1+x} = \frac{1}{2} tan-1 x
⇒ tan-1 1 – tan-1 x = \frac{1}{2} tan-1 x
[∵ tan-1 x – tan-1 y = tan-1 \frac{(x-y)}{1+x y}]
[∵ tan-1 (1) = \frac{\pi}{4}]
⇒ \frac{\pi}{4} = \frac{3}{2} tan-1 x
⇒ tan-1 x = \frac{\pi}{6}
⇒ x = tan \frac{\pi}{6}
∴ x = \frac{1}{\sqrt{3}}.
Question 15.
sin(tan-1 x), |x| < 1 is equal to
(A) \frac{x}{\sqrt{1-x^{2}}}
(B) \frac{1}{\sqrt{1-x^{2}}}
(C) \frac{1}{\sqrt{1+x^{2}}}
(D) \frac{x}{\sqrt{1+x^{2}}}
Solution.
Let tan-1 x = y.
Then, tan y = x
⇒ sin y = \frac{x}{\sqrt{1+x^{2}}}
∴ y = sin-1 (\frac{x}{\sqrt{1+x^{2}}})
⇒ tan-1 x = sin-1 (\frac{x}{\sqrt{1+x^{2}}})
Now, sin(tan-1 x) = sin(sin-1 (\frac{x}{\sqrt{1+x^{2}}}))
= \frac{x}{\sqrt{1+x^{2}}}
The correct answer is (D).
Question 16.
sin-1 (1 – x) – 2 sin-1 x = \frac{\pi}{2}, then x is equal to
(A) 0, \frac{1}{2}
(B) 1, \frac{1}{2}
(C) 0
(D) \frac{1}{2}
Solution.
Given, sin-1 (1 – x) – 2 sin-1 x = \frac{\pi}{2}
putting \frac{\pi}{2} = sin-1 (1 – x) + cos-1 (1 – x)
or sin-1 (1 – x) – 2 sin-1 (1 – x) = sin-1 (1 – x) + cos-1 (1 – x)
⇒ – 2 sin-1 x = cos-1 (1 – x)
Let sin-1 x = α
∴ sin α = x
∴ – 2 sin-1 x = – 2 α = cos-1 (1 – x)
or cos 2α = 1 – x [∵ cos(- θ) = cos θ]
∴ 1 – 2 sin2 α = (1 – x)
Putting sin α = x
⇒ 1 – 2x2 = 1 – x
or 2x2 – x = 0
x(2x – 1) = 0
∴ x = 0, \frac{1}{2}
But x = \frac{1}{2} does not satisfy the equation.
∴ x = 0
Hence, the correct answer is (C).
Question 17.
tan-1 \left(\frac{x}{y}\right) – tan-1 \frac{x-y}{x+y} is equal to
(A) \frac{\pi}{2}
(B) \frac{\pi}{3}
(C) \frac{\pi}{4}
(D) \frac{3 \pi}{4}
Solution.
We have tan-1 \left(\frac{x}{y}\right) – tan-1 \frac{x-y}{x+y}