You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.3

Question 1.

Find the domain of the following functions:

(i) \(\tan ^{-1}(\sqrt{9-x^{2}})\)

(ii) \(\frac{1}{2} \tan ^{-1}\left(1-x^{2}\right)-\frac{\pi}{4}\)

Solution:

(i) tan^{-1} (\(\sqrt {9-x^2}\))

9 – x² ≥ 0 ⇒ 9 ≥ x²

x² ≤ 9 ⇒ x ≤ ± 3

Domain [-3, 3]

Since tan x is an odd function and symmetric about the origin, tan^{-1} x should be an increasing function in its domain.

∴ Domain is (2n + 1)\(\frac{\pi}{2}\)

The domain of y is (-∞, ∞) {x | x ∈ -1} and range is [-1, ∞) {y | y ≥ -1}

The domain for tan^{-1}(x^{2} – 1) is (2n + 1)π. Since tan x is an odd function.

Question 2.

Find the value of

(i) \(\tan ^{-1}\left(\tan \frac{5 \pi}{4}\right)\)

(ii) \(\tan ^{-1}\left(\tan \left(-\frac{\pi}{6}\right)\right)\)

Solution:

Question 3.

Find the value of

(i) \(\tan \left(\tan ^{-1} \frac{7 \pi}{4}\right)\)

(ii) tan(tan^{-1}(1947))

(iii) tan(tan^{-1}(-0.2021))

Solution:

we know that tan(tan^{-1}(x)) = x

(i) tan(tan^{-1}(\(\frac {7π}{4}\))) = \(\frac {7π}{4}\)

(ii) tan(tan^{-1}(1947)) = 1947

(iii) tan(tan^{-1}(-0.2021)) = -0.2021

Question 4.

Find the value of

(i) \(\tan \left(\cos ^{-1}\left(\frac{1}{2}\right)-\sin ^{-1}\left(-\frac{1}{2}\right)\right)\)

(ii) \(\sin \left(\tan ^{-1}\left(\frac{1}{2}\right)-\cos ^{-1}\left(\frac{4}{5}\right)\right)\)

(iii) \(\cos \left(\sin ^{-1}\left(\frac{4}{5}\right)-\tan ^{-1}\left(\frac{3}{4}\right)\right)\)

Solution:

### Samacheer Kalvi 12th Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4.3 Additional Problems

Question 1.

Find the principle value of:

Solution:

Question 2.

Find the value of

Solution:

Question 3.

Find the value of

Solution: