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## 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) f(x) = \(\tan ^{-1}(\sqrt{9-x^{2}})\)

We know the domain of tan^{-1} x is (-∞, ∞) and range is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)

So, the domain of f(x) = \(\tan ^{-1}(\sqrt{9-x^{2}})\) is the set of values of x satisfying the inequality

\(-\infty \leq \sqrt{9-x^{2}} \leq \infty\)

⇒ 9 – x^{2} ≥ 0

⇒ x^{2} ≤ 9

⇒ |x| ≤ 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) \(\tan \left(\tan ^{-1} \frac{7 \pi}{4}\right)=\frac{7 \pi}{4}\)

(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: