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

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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: