ML Aggarwal Class 12 Maths Solutions Section A Chapter 2 Inverse Trigonometric Functions Ex 2.1

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 2 Inverse Trigonometric Functions Ex 2.1

Find the principal values of the following (1 to 6):

Question 1.
(i) sin^-1 (\(\frac{1}{\sqrt{2}}\))
(ii) cos^-1 (\(\frac{1}{2}\))
(iii) tan^-1 (\((\sqrt{3})\))
Solution:
(i) Let sin^-1 (\(\frac{1}{\sqrt{2}}\)) = y, – \(\frac{\pi}{2}\) ≤ y ≤ \(\frac{\pi}{2}\)
sin y = \(\frac{1}{\sqrt{2}}\)
since \(\frac{\pi}{4}\) ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
∴ required principal value = \(\frac{\pi}{4}\)

(ii) Let cos^-1 (\(\frac{1}{2}\)) = y; y ∈ [0, π]
⇒ \(\frac{1}{2}\) = cos y
⇒ cos y = cos \(\frac{\pi}{3}\)
⇒ y = \(\frac{\pi}{3}\) ∈ [0, π]
⇒ cos ^-1 \(\frac{1}{2}\) = \(\frac{\pi}{3}\)

(iii) Let tan^-1 (\((\sqrt{3})\)) = y, – \(\frac{\pi}{2}\) < y < \(\frac{\pi}{2}\)
⇒ tan y = √3 = tan \(\frac{\pi}{3}\)
⇒ y = \(\frac{\pi}{3}\) is the principal value.
[∵ \(\frac{\pi}{3}\) ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))]

Question 2.
(i) sec^-1 (\(\frac{2}{\sqrt{3}}\)) (NCERT)
(ii) cot^-1 (√3) (NCERT)
(iii) cosec^-1 (√2) (NCERT)
Solution:
(i) Let sec^-1 (\(\frac{2}{\sqrt{3}}\)) = y;
y ∈ [0, \(\frac{\pi}{2}\)) ∪ (\(\frac{\pi}{2}\), π]
⇒ \(\frac{2}{\sqrt{3}}\) = sec y
⇒ sec y = sec \(\frac{\pi}{6}\)
⇒ y = \(\frac{\pi}{6}\) ∈ [0, \(\frac{\pi}{2}\)) ∪ (\(\frac{\pi}{2}\), π]
∴ sec^-1 (\(\frac{2}{\sqrt{3}}\)) = \(\frac{\pi}{6}\)

(ii) Let cot^-1 (√3) = y; y ∈ (0, π)
⇒ cot y = √3 = cot \(\frac{\pi}{6}\)
⇒ y = \(\frac{\pi}{6}\) ∈ (0, π)
∴ cot^-1 (√3) = \(\frac{\pi}{6}\)

(iii) Let cosec^-1 (√2) = y;
y ∈ [- \(\frac{\pi}{2}\), 0) ∪ (0, \(\frac{\pi}{2}\)]
⇒ √2 = cosec y
⇒ cosec y = cosec \(\frac{\pi}{4}\)
⇒ y = \(\frac{\pi}{4}\) ∈ [- \(\frac{\pi}{2}\), 0) ∪ (0, \(\frac{\pi}{2}\)]
cosec^-1 (√2) = \(\frac{\pi}{4}\).

Question 3.
(i) cos^-1 (- \(\frac{1}{2}\))
(ii) tan^-1 (- 1)
(iii) cosec^-1 (- √2)
Solution:
(i) Let cos^-1 (- \(\frac{1}{2}\)) = y, 0 ≤ y ≤ π
⇒ cos y = – \(\frac{1}{2}\)
= – cos \(\frac{\pi}{3}\)
= cos (π – \(\frac{\pi}{2}\))
⇒ y = \(\frac{2 \pi}{3}\) ∈ [0, π]
∴ Required principal value = \(\frac{2 \pi}{3}\).

(ii) Let tan^-1 (- 1) = y, – \(\frac{\pi}{2}\) < y < \(\frac{\pi}{2}\)
⇒ tan y = – 1
= – tan \(\frac{\pi}{4}\)
= tan (- \(\frac{\pi}{4}\))
⇒ y = – \(\frac{\pi}{4}\) is the required principal value.
[∵ – \(\frac{\pi}{4}\) ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)).

(iii) Let cosec^-1 (- √2) = y
y ∈ [- \(\frac{\pi}{2}\), 0) ∪ (0, \(\frac{\pi}{2}\)]
⇒ cosec y = – √2
= – cosec \(\frac{\pi}{4}\)
= cosec (- \(\frac{\pi}{4}\))
∴ y = – \(\frac{\pi}{4}\) ∈ [- \(\frac{\pi}{2}\), 0) ∪ (0, \(\frac{\pi}{2}\)]
⇒ cosec^-1 (- √2) = – \(\frac{\pi}{4}\)

Question 4.
(i) cot^-1 (- \(\frac{1}{\sqrt{3}}\))
(ii) sec^-1 (- \(\frac{2}{\sqrt{3}}\))
(iii) sin^-1 (- 1)
Solution:
(i) Let cot^-1 (- \(\frac{1}{\sqrt{3}}\)) = y; y ∈ (0, π)
⇒ cot y = – \(\frac{1}{\sqrt{3}}\)
= – cot (\(\frac{\pi}{3}\))
= cot (π – \(\frac{\pi}{3}\))
⇒ cot y = cot \(\frac{2 \pi}{3}\)
⇒ y = \(\frac{2 \pi}{3}\) ∈ (0, π)
∴ cot^-1 (- \(\frac{1}{\sqrt{3}}\)) = \(\frac{2 \pi}{3}\)

(ii) Let sec^-1 (- \(\frac{2}{\sqrt{3}}\)) = y;
y ∈ [0, \(\frac{\pi}{2}\)) ∪ (\(\frac{\pi}{2}\), π]
⇒ sec y = – \(\frac{2}{\sqrt{3}}\)
= – sec \(\frac{\pi}{6}\)
⇒ sec y = sec (π – \(\frac{\pi}{6}\))
= sec \(\frac{5 \pi}{6}\)
⇒ y = \(\frac{5 \pi}{6}\) ∈ [0, \(\frac{\pi}{2}\)) ∪ (\(\frac{\pi}{2}\), π]
∴ sec^-1 (- \(\frac{2}{\sqrt{3}}\)) = \(\frac{5 \pi}{6}\)

(iii) Let sin^-1 (- 1) = y ; y ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
⇒ sin y = – 1
= – sin \(\frac{\pi}{2}\)
⇒ sin y = sin (- \(\frac{\pi}{2}\))
⇒ y = – \(\frac{\pi}{2}\) ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
∴ sin^-1 (- 1) = – \(\frac{\pi}{2}\)

Question 5.
(i) cos^-1 (cos \(\frac{4 \pi}{3}\))
(ii) sin^-1 (sin \(\frac{2 \pi}{3}\))
(iii) tan^-1 (tan \(\frac{3 \pi}{4}\))
Solution:
(i) cos^-1 (cos \(\frac{4 \pi}{3}\)) ≠ \(\frac{4 \pi}{3}\)
(∵ \(\frac{4 \pi}{3}\) ∉ [0, π]
∴ cos^-1 (cos \(\frac{4 \pi}{3}\)) = cos^-1 {cos (2π – 4π/3)}
= cos^-1 (cos 2π/3)
[∵ cos (2π – θ) = cos θ]
= cos \(\frac{2 \pi}{3}\)
[∵ cos^-1 (cos θ) = θ if θ ∈ [0, π]

(ii) sin^-1 (sin \(\frac{2 \pi}{3}\)) ≠ \(\frac{2 \pi}{3}\)
[∵ sin^-1 (sin x) = x ∀ x ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]]
sin^-1 (sin \(\frac{2 \pi}{3}\)) = sin^-1 {sin (π – \(\frac{\pi}{3}\))}
= sin^-1 (sin \(\frac{\pi}{3}\)) = \(\frac{\pi}{3}\)
[∵ \(\frac{\pi}{3}\) ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]]

(iii) tan^-1 (tan \(\frac{3 \pi}{4}\)) = \(\frac{3 \pi}{4}\) ∉ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
which is the principal value branch of tan^-1
Now tan^-1 (tan \(\frac{3 \pi}{4}\)) = tan^-1 (tan (π – \(\frac{\pi}{4}\)))
= tan^-1 (- tan \(\frac{\pi}{4}\))
= tan^-1 (tan (\(\frac{\pi}{4}\)))
= – \(\frac{\pi}{4}\) ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))

Question 6.
(i) cos^-1 (cos \(\frac{5 \pi}{3}\))
(ii) tan^-1 (tan \(\frac{9 \pi}{8}\))
(iii) sec^-1 (sec \(\frac{9 \pi}{8}\))
Solution:
(i) cos^-1 (cos \(\frac{5 \pi}{3}\)) ≠ \(\frac{5 \pi}{3}\)
∵ \(\frac{5 \pi}{3}\) ∉ [0, π]
∴ cos^-1 (cos \(\frac{5 \pi}{3}\)) = cos^-1 {cos (2π – \(\frac{\pi}{3}\))}
= cos^-1 (cos \(\frac{\pi}{3}\))
[∵ cos (2π – θ) = cos θ]
= \(\frac{\pi}{3}\)
[∵ cos^-1 (cos θ) = θ ∈ θ [0, π]]

(ii) tan^-1 (tan \(\frac{9 \pi}{8}\)) ≠ \(\frac{9 \pi}{8}\)
Since \(\frac{9 \pi}{8}\) ∉ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
Now tan^-1 (tan \(\frac{\pi}{8}\))
[∵ tan (π + θ) = tan θ]
= \(\frac{\pi}{8}\), Since \(\frac{\pi}{8}\) ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
[∵ tan^-1 (tan θ) = θ ∀ θ ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))]

(iii) sec^-1 (sec \(\frac{9 \pi}{5}\)) ≠ \(\frac{9 \pi}{5}\)
∵ \(\frac{9 \pi}{5}\) ∉ [0, \(\frac{\pi}{2}\)) ∪ (\(\frac{\pi}{2}\), π]
∴ sec^-1 (sec \(\frac{9 \pi}{5}\)) = sec^-1 (sec (2π – \(\frac{9 \pi}{5}\)))
= sec^-1 (sec \(\frac{\pi}{5}\))
= \(\frac{\pi}{5}\)
[∵ sec (2π – θ) = sec θ]
[∵ sec^-1 (sec θ) = θ ∀ θ ∈ [0, \(\frac{\pi}{2}\)) ∪ (\(\frac{\pi}{2}\), π]]

Question 7.
Show that:
(i) tan^-1 (tan \(\frac{5 \pi}{6}\)) ≠ \(\frac{5 \pi}{6}\), what is its value?
(ii) cos^-1 (cos (- \(\frac{\pi}{6}\))) ≠ \(-\frac{\pi}{6}\), what is its value?
(iii) sin^-1 (sin \(\frac{5 \pi}{3}\)) ≠ \(\frac{5 \pi}{3}\), what is its value?
Solution:
(i) Now, tan^-1 (tan \(\frac{5 \pi}{6}\)) ≠ \(\frac{5 \pi}{6}\)
∴ \(\frac{5 \pi}{6}\) ∉ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
∴ tan^-1 (tan \(\frac{5 \pi}{6}\)) = tan^-1 {tan (π – \(\frac{\pi}{6}\))}
= tan^-1 (- tan \(\frac{\pi}{6}\))
[∵ tan (π – θ) = – tan θ]
= tan^-1 {tan (- \(\frac{\pi}{6}\))}
= – \(\frac{\pi}{6}\) ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
[∵ tan^-1 (tan x) = x ∀ x ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))]

(ii) cos^-1 (cos (- \(\frac{\pi}{6}\))) ≠ \(-\frac{\pi}{6}\)
∵ \(-\frac{\pi}{6}\) ∉ [0, π]
Now, cos^-1 (cos (- \(\frac{\pi}{6}\))) = cos^-1 (cos \(\frac{\pi}{6}\))
[∵ cos (- θ) = cos θ]
= \(\frac{\pi}{6}\)
[∵ cos^-1 (cos θ) = θ ∀ θ ∈ [0, π] and \(\frac{\pi}{6}\) ∈ [0, π]]

(iii) sin^-1 (sin \(\frac{5 \pi}{3}\)) ≠ \(\frac{5 \pi}{3}\)
∵ \(\frac{5 \pi}{3}\) ∉ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
Now sin^-1 (sin \(\frac{5 \pi}{3}\)) = sin^-1 (sin (2π – \(\frac{\pi}{3}\)))
= sin^-1 (- sin \(\frac{\pi}{3}\))
[∵ sin (2π – θ) = – sin θ]
= sin^-1 (sin (\(\frac{\pi}{3}\)))
= – \(\frac{\pi}{3}\)
[∵ sin^-1 (sin θ) = θ ∀ θ ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]

Question 8.
Find the domain of the following functions:
(i) sin^-1 (1 – x)
(ii) sec^-1 (2x – 3)
Solution:
(i) Given f(x) sin^-1 (1 – x)
The domain of sin^-1 x be [- 1, 1]
∴ f(x) is defined for all x sa1isting – 1 ≤ 1 – x ≤ 1
⇒ – 2 ≤ – x ≤ 0
⇒ 2 ≥ x ≥ 0
⇒ 0 ≤ x ≤ 2
⇒ x ∈ [0, 2]
∴ domain of f(x) = sin^-1 (1 – x) be [0, 2]

(ii) Given f(x) = sec^-1 (2x – 3)
The domain of sec^-1 x be (- ∞, – 1] ∪ [1, ∞)
∴ f(x) is defined for all x satisfying
|2x – 3| ≥ + 1
⇒ 2x – 3 ≥ 1 or
⇒ 2x – 3 ≤ – 1
⇒ 2x ≥ 4 or 2x ≤ 2
⇒ x ≥ 2 or x ≤ 1
∴ x ∈ (- ∞, 1] ∪ [2, ∞)
Thus domain of f(x) be (- ∞, 1] ∪ {2, ∞).

Question 9.
Write the range of one branch of sin^-1 x, other than the principal branch.
Solution:
If we have restricted the domain of sine function to any one of the intervals i.e.
\(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right],\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right],\left[\frac{3 \pi}{2}, \frac{5 \pi}{2}\right]\) i.e.
in general [nπ – \(\frac{\pi}{2}\), nπ + \(\frac{\pi}{2}\)] then inverse of sine function
i.e. sin^-1 x is exists in any of these intervals with domain [- 1, 1] and range \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right],\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]\) and so on.
Hence range of one branch of sin^-1 x other than principal branch be \(\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]\).

Question 10.
Write the range of one branch of cos^-1 x, other thait the principal branch.
Solution:
If we have restricted the domain of cosine function to any one of the intervals i.e. [- π, 0], [0, π], [π, 2π] i.e.
in general [nπ, (n + 1) π] then inverse of cosine function
i.e. cos^-1 x exists in any one of these intervals with domain [- 1, 1] and range [- π, 0] or [0, π] or [π, 2π] and so on.
Hence, rante of one branch of cos^-1 x other than principal branch be [π, 2π].

Question 11.
II cot^-1 (\(\frac{3}{4}\)) = x, find the value of cos x.
Solution:
Given cot^-1 (\(\frac{3}{4}\)) = x
⇒ cot x = \(\frac{3}{4}\); x ∈ (0, π)
For these values of x, sin x > 0

∴ sin x = \(\frac{4}{5}\)
Thus, cos x = cot x. sin x
= \(\frac{3}{4} \times \frac{4}{5}\)
cos x = \(\frac{3}{5}\)

Question 12.
(i) If tan^-1 (\(\frac{3}{4}\)) = x, find the values of cos x and sinx.
(ii) If cot^-1 (- \(\frac{1}{7}\)) = x, find the values of sin x and cos x.
(iii) If tan^-1 x = sin^-1 (\(\frac{1}{2}\)), find the value of x.
Solution:
(i) Given tan^-1 \(\frac{3}{4}\) = x
⇒ tan x = \(\frac{3}{4}\); – \(\frac{\pi}{2}\) < x < \(\frac{\pi}{2}\)
∴ sec x = \(\sqrt{1+\tan ^2 x}\)
= \(\sqrt{1+\frac{9}{16}}=\frac{5}{4}\)
∴ cos x = \(\frac{4}{5}\)
now sin x = tan x × cos x
= \(\frac{3}{4} \times \frac{4}{5}=\frac{3}{5}\)

(ii) Given cot^-1 (- \(\frac{1}{7}\)) = x
⇒ cot x = – \(\frac{1}{7}\); x ∈ (0, π)
For these values of x, sin x, cosec x > 0
∴ cosec x = \(\sqrt{1+\cot ^2 x}\)
= \(\sqrt{1+\frac{1}{49}}\)
= \(\frac{\sqrt{50}}{7}=\frac{5 \sqrt{2}}{7}\)
⇒ sin x = \(\frac{7}{5 \sqrt{2}}\)
∴ cos x = cot x . sin x
= \(-\frac{1}{7} \times \frac{7}{5 \sqrt{2}}=-\frac{1}{5 \sqrt{2}}\)

(iii) Given tan^-1 x = sin^-1 \(\frac{1}{2}\) = \(\frac{\pi}{6}\)
[∵ \(\frac{\pi}{6}\) ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]]
⇒ x = tan (\(\frac{\pi}{6}\)) = \(\frac{1}{\sqrt{3}}\)
[Also \(\frac{\pi}{6}\) ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]]

Question 13.
Evaluate the following:
(i) cot (tan^-1 √3)
(ii) sin (\(\frac{\pi}{6}\) – sin^-1 (- \(\frac{\sqrt{3}}{2}\)))
(iii) cos (cos^-1 (- \(\frac{\sqrt{3}}{2}\)) + \(\frac{\pi}{6}\)) (NCERT Examplar)
Solution:
(i) Let tan^-1 √3 = y;
y ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
⇒ tan y = √3 = tan \(\frac{\pi}{3}\)
⇒ y = \(\frac{\pi}{3}\) ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))

(ii) Let sin^-1 (- \(\frac{\sqrt{3}}{2}\)) = y;
y ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
⇒ sin y = – \(\frac{\sqrt{3}}{2}\)
= – sin \(\frac{\pi}{3}\)
⇒ sin y = sin (- \(\frac{\pi}{3}\))
⇒ y = – \(\frac{\pi}{3}\) ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
∴ sin [\(\frac{\pi}{6}\) – sin^-1 (- \(\frac{\sqrt{3}}{2}\))]
= sin [\(\frac{\pi}{6}\) – (- \(\frac{\pi}{3}\))]
= sin (\(\frac{\pi}{2}\)) = 1

(iii) Let cos^-1 (- \(\frac{\sqrt{3}}{2}\)) = y; y ∈ [0, π]
⇒ cos y = – \(\frac{\sqrt{3}}{2}\)
= – cos \(\frac{\pi}{6}\)
⇒ cos y = cos (π – \(\frac{\pi}{6}\))
= cos \(\frac{5 \pi}{6}\)
⇒ y = \(\frac{5 \pi}{6}\) [∵ \(\frac{5 \pi}{6}\) ∈ [0, π]]
∴ cos^-1 (- \(\frac{\sqrt{3}}{2}\)) = \(\frac{5 \pi}{6}\)
Thus, cos (cos^-1 (- \(\frac{\sqrt{3}}{2}\)) + \(\frac{\pi}{6}\)
= cos (\(\frac{5 \pi}{6}\) + \(\frac{\pi}{6}\))
= cos π = – 1

Question 14.
Prove the following:
(i) sin^-1 (- \(\frac{1}{2}\)) + cos^-1 (- \(\frac{\sqrt{3}}{2}\)) = \(\frac{2 \pi}{3}\)
(ii) tan^-1 (- 1) + cos^-1 (- \(\frac{1}{\sqrt{2}}\)) = \(\frac{\pi}{2}\)
Solution:
(i) Let sin^-1 (- \(\frac{1}{2}\)) = y;
y ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
⇒ sin y = – \(\frac{1}{2}\)
= – sin \(\frac{\pi}{6}\)
= sin (- \(\frac{\pi}{6}\))
⇒ y = – \(\frac{\pi}{6}\) ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
∴ sin^-1 (- \(\frac{1}{2}\)) = – \(\frac{\pi}{6}\) ………….(1)
Let cos^-1 (- \(\frac{\sqrt{3}}{2}\)) = Z; Z ∈ [0, π]
⇒ cos Z = – \(\frac{\sqrt{3}}{2}\)
= – cos \(\frac{\pi}{6}\)
⇒ cos Z = cos (π – \(\frac{\pi}{6}\))
⇒ cos Z = cos \(\frac{5 \pi}{6}\)
⇒ Z = \(\frac{5 \pi}{6}\) ∈ [0, π]
∴ sin^-1 (- \(\frac{1}{2}\)) + cos^-1 (- \(\frac{\sqrt{3}}{2}\))
= \(-\frac{\pi}{6}+\frac{5 \pi}{6}\) [using (1) and (2)]
= \(\frac{4 \pi}{6}=\frac{2 \pi}{3}\)

(ii) Let tan^-1 (- 1) = y;
y ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
⇒ tan y = – 1
= – tan \(\frac{\pi}{4}\)
= tan (- \(\frac{\pi}{4}\))
⇒ y = – \(\frac{\pi}{4}\) ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
∴ tan^-1 (- 1) = – \(\frac{\pi}{4}\) ……………(1)
Let cos^-1 (- \(\frac{1}{\sqrt{2}}\)) = Z;
Z ∈ [0, π]
⇒ cos Z = – \(\frac{1}{\sqrt{2}}\)
= – cos \(\frac{\pi}{4}\)
⇒ cos Z = cos [π – \(\frac{\pi}{4}\)]
= cos \(\frac{3 \pi}{4}\)
⇒ Z = \(\frac{3 \pi}{4}\) ∈ [0, π]
∴ cos^-1 (- \(\frac{1}{\sqrt{2}}\)) = \(\frac{3 \pi}{4}\) …………(2)
∴ tan^-1 (- 1) + cos^-1 (- \(\frac{1}{\sqrt{2}}\))
= \(-\frac{\pi}{4}+\frac{3 \pi}{4}=\frac{\pi}{2}\) [using (1) and (2)]

Question 15.
Using principal values, find the values of:
(i) cos^-1 (\(\frac{1}{2}\))- 2 sin^-1(- \(\frac{1}{2}\))
(ii) tan^-1 (√3) – cot^-1 (- √3)
(iii) tan^-1 (1) + cos^-1 (- \(\frac{1}{2}\))
(iv) cosec^-1 (- 1) + cot^-1 (- \(\frac{1}{\sqrt{3}}\))
Solution:
(i) Let cos^-1 (\(\frac{1}{2}\)) = y ; y ∈ [0, π]
⇒ cos y = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\)
⇒ y = \(\frac{\pi}{3}\)
⇒ cos^-1 \(\frac{1}{2}\) = \(\frac{\pi}{3}\) ……….(1)
Let sin^-1 (- \(\frac{1}{2}\)) = z;
z ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
⇒ sin z = – \(\frac{1}{2}\)
= – sin \(\frac{\pi}{6}\)
= sin (- \(\frac{\pi}{6}\))
⇒ z = – \(\frac{\pi}{6}\) ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
⇒ sin^-1 (- \(\frac{1}{2}\)) = – \(\frac{\pi}{6}\) ……………..(2)
Thus cos^-1 (\(\frac{1}{2}\)) – 2 sin^-1 (- \(\frac{1}{2}\))
= \(\frac{\pi}{3}-2 \times\left(-\frac{\pi}{6}\right)=\frac{2 \pi}{3}\) [using (1) and (2)]

(ii) Let tan^-1 √3 = y ;
y ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
⇒ tan y = √3 = tan \(\frac{\pi}{3}\)
∴ y = \(\frac{\pi}{3}\) ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
∴ tan^-1 √3 = \(\frac{\pi}{3}\) ……………..(1)
Let cot^-1 (- √3) = Z ; Z ∈ [0, π]
⇒ cot Z = – √3 = – cot (\(\frac{\pi}{6}\))
⇒ cot Z = cot (π – \(\frac{\pi}{6}\))
⇒ Z = π – \(\frac{\pi}{6}\)
= \(\frac{5 \pi}{6}\) ∈ [0, π]
∴ cot^-1 (- √3) = \(\frac{5 \pi}{6}\) ……….(2)
Thus tan (√3) – cot^-1 (- √3)
= \(\frac{\pi}{3}-\frac{5 \pi}{6}=-\frac{\pi}{2}\) [using (1) and (2)]

(iii) Let tan^-1 (1) = y ;
y ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
⇒ tan y = 1 = tan \(\frac{\pi}{4}\)
⇒ y = \(\frac{\pi}{4}\) ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
∴ tan^-1 (1) = \(\frac{\pi}{4}\) ……………(1)
Let cos^-1 (- \(\frac{1}{2}\)) = Z ; Z ∈ [0, π]
⇒ cos Z = – \(\frac{1}{2}\)
= – cos \(\frac{\pi}{3}\)
⇒ cos Z = cos (π – \(\frac{\pi}{3}\))
⇒ Z = π – \(\frac{\pi}{3}\) = \(\frac{2 \pi}{3}\) ∈ [0, π]
∴ cos^-1 (- \(\frac{1}{2}\)) = \(\frac{2 \pi}{3}\) ………….(2)
Thus tan^-1 (1) + cos^-1 (- \(\frac{1}{2}\)) = \(\frac{\pi}{4}+\frac{2 \pi}{3}\) [using (1) and (2)]
= \(\frac{11 \pi}{12}\)

(iv) Let cosec^-1 (- 1) = y;
y ∈ [- \(\frac{\pi}{2}\), 0) ∪ (0, \(\frac{\pi}{2}\)]
∴ cosec^-1 (- 1) = – \(\frac{\pi}{2}\) …………….(1)
Let cot^-1 (- \(\frac{1}{\sqrt{3}}\)) = Z ;
Z ∈ [0, π]
∴ cot Z = – \(\frac{1}{\sqrt{3}}\)
= – cot \(\frac{\pi}{3}\)
⇒ cot Z = cot (π – \(\frac{\pi}{3}\))
= cot \(\frac{2 \pi}{3}\)
⇒ Z = \(\frac{2 \pi}{3}\) ∈ [0, π]
∴ cot^-1 (- \(\frac{1}{\sqrt{3}}\)) = \(\frac{2 \pi}{3}\) …………..(2)
Thus, cosec^-1 (- 1) + cot^-1 (- \(\frac{1}{\sqrt{3}}\))
= \(-\frac{\pi}{2}+\frac{2 \pi}{3}=\frac{\pi}{6}\)

Question 16.
Find the values of the following:
(i) tan^-1 (2 cos (2 sin^-1 \(\frac{1}{2}\)))
(ii) tan^-1 (2 sin (2 cos^-1 \(\frac{\sqrt{3}}{2}\)))
(iii) sin^-1 (\(\frac{1}{2}\) sin (2 cot^-1 (- 1)))
(iv) cot (sin^-1 (cos (tan^-1 1)))
Solution:
(i) tan^-1 (2 cos (2 sin^-1 \(\frac{1}{2}\)))
= tan^-1 {2 cos (2 × \(\frac{\pi}{2}\))}
[∵ sin^-1 \(\frac{1}{2}\) = \(\frac{\pi}{6}\) ∈ [- \(\frac{\pi}{2}\), 0) ∪ (0, \(\frac{\pi}{2}\)]]
= tan^-1 {2 cos \(\frac{\pi}{3}\)}
= tan^-1 {2 × \(\frac{1}{2}\)} = tan^-1 (1)
= \(\frac{\pi}{4} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)

(ii) We know that cos^-1 \(\frac{\sqrt{3}}{2}\) = \(\frac{\pi}{6}\)
∴ tan^-1 {2 sin (4 cos^-1 \(\frac{\sqrt{3}}{2}\) )}
= tan^-1 {2 sin (4 × \(\frac{\pi}{6}\))}
= tan^-1 {2 sin \(\frac{2 \pi}{3}\)}
= tan^-1 {2 sin (π – \(\frac{\pi}{3}\))}
= tan^-1 {2 sin \(\frac{\pi}{3}\)}
= tan^-1 {2 × \(\frac{\sqrt{3}}{2}\)}
= tan^-1 (√3)
= \(\frac{\pi}{3}\) ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))

(iii) Let cot^-1 (- 1) = x ; 0 < x < π
⇒ cot x = – 1 ; 0 < x < π
⇒ tan x = – 1 = – tan \(\frac{\pi}{4}\)
= tan (π – \(\frac{\pi}{4}\))
= tan \(\frac{3 \pi}{4}\)
⇒ x = \(\frac{3 \pi}{4}\) ∈ (0, π)
∴ sin^-1 (\(\frac{1}{2}\) sin (2 cot^-1 (- 1)))

(iv) cot (sin^-1 (cos (tan^-1 1)))
= cot (sin^-1 (cos \(\frac{\pi}{4}\)))
= cot (sin^-1 \(\frac{1}{\sqrt{2}}\))
= cot (sin^-1 (sin \(\frac{\pi}{4}\)))
= cot \(\frac{\pi}{4}\))) = 1
[∵ sin^-1 (sin x) = x ∀ x ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]]

Question 17.
Find the value of:
(i) tan^-1 (tan \(\frac{5 \pi}{6}\)) + cos^-1 (cos \(\frac{13 \pi}{6}\)) (NCERT Exemplar)
(ii) tan^-1 (- \(\frac{1}{\sqrt{3}}\)) + cot^-1 (\(\frac{1}{\sqrt{3}}\)) + tan^-1 (sin (- \(\frac{\pi}{2}\)))
Solution:
(i) Since \(\frac{5 \pi}{6}\) ∉ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
Now tan^-1 (tan \(\frac{5 \pi}{6}\)) = tan^-1 [tan (π – \(\frac{\pi}{6}\))]
= tan^-1 {- tan \(\frac{\pi}{6}\)}
= tan^-1 (- \(\frac{1}{\sqrt{3}}\)) …………(1)
Let tan^-1 (- \(\frac{1}{\sqrt{3}}\)) = y
⇒ tan y = – \(\frac{1}{\sqrt{3}}\)
= – tan \(\frac{\pi}{6}\)
= tan (- \(\frac{\pi}{6}\))
∴ y = – \(\frac{\pi}{6}\) ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
Thus, tan^-1 (- \(\frac{1}{\sqrt{3}}\)) = – \(\frac{\pi}{6}\)
∴ from (1) ; tan^-1 (tan \(\frac{5 \pi}{6}\)) = – \(\frac{\pi}{6}\) ………..(2)
Now cos^-1 {cos (\(\frac{13 \pi}{6}\))}
= cos^-1 {cos (2π + \(\frac{\pi}{6}\))} [Since \(\frac{13 \pi}{6}\) ∉ [0, π]]
= cos^-1 {cos \(\frac{\pi}{6}\)}
= cos^-1 \(\left(\frac{\sqrt{3}}{2}\right)\)
= \(\frac{\pi}{6}\) ……………..(3)
Thus tan^-1 (tan \(\frac{5 \pi}{6}\)) + cos^-1 (cos \(\frac{13 \pi}{6}\))
= \(-\frac{\pi}{6}+\frac{\pi}{6}\) = 0 [Using (2) and (3)]

(ii) Let tan^-1 (- \(\frac{1}{\sqrt{3}}\)) = x ;
x ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
⇒ tan x = – \(\frac{1}{\sqrt{3}}\)
= – tan \(\frac{\pi}{6}\)
= tan (- \(\frac{\pi}{6}\))
∴ x = – \(\frac{\pi}{6}\) ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
⇒ tan^-1 (- \(\frac{1}{\sqrt{3}}\)) = – \(\frac{\pi}{6}\) …………..(1)
Let cot^-1 (\(\frac{1}{\sqrt{3}}\)) = y ;
y ∈ (0, π)
⇒ cot y = \(\frac{1}{\sqrt{3}}\) = cot \(\frac{\pi}{3}\)
⇒ y = \(\frac{\pi}{3}\) ∈ (0, π)
⇒ cot^-1 (\(\frac{1}{\sqrt{3}}\)) = \(\frac{\pi}{3}\) …………..(2)
∴ tan^-1 {sin (- \(\frac{\pi}{2}\))} = tan^-1 {- sin (\(\frac{\pi}{2}\))}
= tan^-1 {- 1} ……………..(3)
Let tan^-1 (- 1) = z
⇒ tan z = – 1
= – tan (\(\frac{\pi}{4}\))
= tan (- \(\frac{\pi}{4}\))
⇒ z = – \(\frac{\pi}{4}\) ∈ (- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\))
∴ from (3) ;
tan^-1 {sin (- \(\frac{\pi}{2}\))} = – \(\frac{\pi}{4}\) ……………..(4)
Thus,
tan^-1 (- \(\frac{1}{\sqrt{3}}\)) + cot^-1 (\(\frac{1}{\sqrt{3}}\)) + tan^-1 (sin (- \(\frac{\pi}{2}\)))
= \(-\frac{\pi}{6}+\frac{\pi}{3}-\frac{\pi}{4}\)
= \(\frac{-2 \pi+4 \pi-3 \pi}{12}=-\frac{\pi}{12}\).

Question 18.
Evaluate the following:
(i) cos (sin^-1 (- \(\frac{3}{5}\)))
(ii) cosec (cos^-1 (- \(\frac{12}{13}\)))
(iii) sin (2 cot^-1 \(\frac{1}{5}\))
Solution:
(i) cos (sin^-1 (- \(\frac{3}{5}\))) = cos (- sin^-1 (- \(\frac{3}{5}\)))
[∵ sin^-1 (- x) = – sin^-1 x ∀ x ∀∈ [- 1, 1]]
= cos (sin^-1 (\(\frac{3}{5}\)))
[∵ cos (- θ) = cos θ]
Now we convert sin^-1 to cos^-1, for this we make a right angled triangle with perpendicular p = 3 and hypotenuse h = 5
∴ base b = \(\sqrt{h^2-p^2}\)
= \(\sqrt{25-9}\) = 4

∴ sin^-1 (\(\frac{3}{5}\)) = cos^-1 (\(\frac{4}{5}\))
Thus, cos (sin^-1 (- \(\frac{3}{5}\)) = cos (sin^-1 (\(\frac{3}{5}\)))
= cos (cos^-1 (\(\frac{4}{5}\)))
= \(\frac{4}{5}\)
[∵ cos (cos^-1 x) = x ∀ x ∈ [- 1, 1]]

(ii) cosec (cos^-1 (- \(\frac{12}{13}\)))
= cosec (cos^-1 (- \(\frac{12}{13}\)))
∴ cos^-1 (\(\frac{12}{13}\)) = cosec^-1 (\(\frac{13}{5}\))
∴ cosec (cos^-1 (- \(\frac{12}{13}\))) = cosec (cosec^-1 (\(\frac{13}{5}\))=
= \(\frac{13}{5}\))
[∵ cosec (cosec^-1 x) = x ∀ | x | ≥ 1]

(iii) put cot^-1 \(\frac{1}{5}\) = θ
⇒ cot θ = \(\frac{1}{5}\)
⇒ tan θ = 5
∴ sin (2 cot^-1 \(\frac{1}{5}\)) = sin 2θ
= \(\frac{2 \tan \theta}{1+\tan ^2 \theta}\)
= \(\frac{2 \times 5}{1+25}\)
= \(\frac{10}{26}=\frac{5}{13}\)

Question 19.
Show that : tan (\(\frac{1}{2}\) sin^-1 \(\frac{3}{4}\)) = \(\frac{4-\sqrt{7}}{2}\).
Solution:
put sin^-1 \(\frac{3}{4}\) = θ
⇒ sin θ = \(\frac{3}{4}\) ;
θ ∈ [- \(\frac{\pi}{2}\), \(\frac{\pi}{2}\)]
∴ tan^-1 [\(\frac{1}{2}\) sin^-1 \(\frac{3}{4}\)] = tan \(\frac{\theta}{2}\)
We know that,
sin θ = \(\frac{2 \tan \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}\)
⇒ \(\frac{3}{4}=\frac{2 \tan \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}\)
⇒ 3 + 3 tan² \(\frac{\theta}{2}\) = 8 tan \(\frac{\theta}{2}\)
⇒ 3 tan² \(\frac{\theta}{2}\) – 8 tan \(\frac{\theta}{2}\) + 3 = 0
⇒ tan \(\frac{\theta}{2}\) = \(\frac{4 \pm \sqrt{7}}{3}\)
[∵ θ ∈ \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)]
⇒ \(\frac{\theta}{2} \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]\)
⇒ – 1 ≤ tan \(\frac{\theta}{2}\) ≤ 1
∴ tan \(\frac{\theta}{2}\) ≠ \(\frac{4+\sqrt{7}}{3}\)]
∴ tan \(\frac{\theta}{2}\) = \(\frac{4-\sqrt{7}}{3}\)