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## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.5

Question 1.
Find the value, if it exists. If not, give the reason for non-existence.
(i) sin-1(cos π)
(ii) $$\tan ^{-1}\left(\sin \left(-\frac{5 \pi}{2}\right)\right)$$
(iii) sin-1[sin 5]
Solution:
(i) sin-1(cos π) = sin-1(-1) = $$-\frac{\pi}{2}$$
(ii) $$\tan ^{-1}\left(\sin \frac{5 \pi}{2}\right)=\tan ^{-1}\left(-\sin \frac{\pi}{2}\right)=\tan ^{-1}(-1)=-\frac{\pi}{4}$$
(iii) sin-1(sin 5) = sin-1[sin (5 – 2π)] = 5 – 2π

Question 2.
Find the value of the expression in terms of x, with the help of a reference triangle.
(i) sin(cos-1(1 – x))
(ii) cos(tan-1(3x – 1))
(iii) $$\tan \left(\sin ^{-1}\left(x+\frac{1}{2}\right)\right)$$
Solution:
(i) Let cos-1(1 – x) = θ
1 – x = cos θ
We know sin2 θ = 1 – cos2 θ

Question 3.
Find the value of
(i) $$\sin ^{-1}\left(\cos \left(\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)\right)\right.$$
(ii) $$\cot \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{4}{5}\right)$$
(iii) $$\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)$$
Solution:
(i) $$\sin ^{-1} \frac{\sqrt{3}}{2}=\frac{\pi}{3} \text { and } \cos \left(\frac{\pi}{3}\right)=\frac{1}{2} \text { and }$$

Question 4.
Prove that
(i) $$\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}$$
(ii) $$\sin ^{-1} \frac{3}{5}-\cos ^{-1} \frac{12}{13}=\sin ^{-1} \frac{16}{65}$$
Solution:

Question 5.
Prove that tan-1 x + tan-1 y + tan-1 z = tan-1 $$\left[\frac{x+y+z-x y z}{1-x y-y z-z x}\right]$$
Solution:

Question 6.
If tan-1 x + tan-1 y + tan-1 z = π, show that x + y + z = xyz.
Solution:
Given tan-1 x + tan-1 y + tan-1 z = π

Question 7.
Prove that $$\tan ^{-1} x+\tan ^{-1} \frac{2 x}{1-x^{2}}=\tan ^{-1} \frac{3 x-x^{3}}{1-3 x^{2}},|x|<\frac{1}{\sqrt{3}}$$
Solution:

Question 8.
Simplify: $$\tan ^{-1} \frac{x}{y}-\tan ^{-1} \frac{x-y}{x+y}$$
Solution:

Question 9.
Find the value of
(i) $$\sin ^{-1} \frac{5}{x}+\sin ^{-1} \frac{12}{x}=\frac{\pi}{2}$$
(ii) $$2 \tan ^{-1} x=\cos ^{-1} \frac{1-a^{2}}{1+a^{2}}-\cos ^{-1} \frac{1-b^{2}}{1+b^{2}}$$, a > 0, b > 0
(iii) 2 tan-1(cos x) = tarn-1 (2 cosec x)
(iv) cot-1 x – cot-1 (x + 2) = $$\frac{\pi}{12}$$, x > 0
Solution:

Question 10.
Find the number of solution of the equation tan-1(x – 1) + tan-1 x + tan-1 (x + 1) = tan-1(3x).
Solution:
tan-1(x – 1) + tan-1 x + tan-1 (x + 1) = tan-1(3x)
tan-1(x – 1) + tan-1 (x + 1) = tan-1 3x – tan-1 x

LHS = RHS
⇒ $$\tan ^{-1} \frac{2 x}{2-x^{2}}=\tan ^{-1} \frac{2 x}{1+3 x^{2}}$$
⇒ $$\frac{2 x}{2-x^{2}}=\frac{2 x}{1+3 x^{2}}$$
⇒ 2 – x2 = 1 + 3x2
⇒ 4x2 = 1
⇒ x2 = $$\frac{1}{4}$$
⇒ x = ±$$\frac{1}{2}$$
So, the equation has 2 solutions.

### Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.5 Additional Questions

Question 1.
Solve the following equation: sin-1(1 – x) – 2 sin-1 x = $$\frac{\pi}{2}$$
Solution:

Question 2.
Solve: tan-1 2x + tan-1 3x = $$\frac{\pi}{4}$$
Solution:

Question 3.

Solution:
Do it yourself

Question 4.

Solution:
Do it yourself

Question 5.

Solution:
Do it yourself

Question 6.

Solution:
Do it yourself

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution: