Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 4 Inverse Trigonometric Functions Ex 4.3 Textbook Questions and Answers, Notes.

## 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} (1 – x²) – \(\frac {π}{4}\)

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

(ii) Range of tan^{-1} x is R

-∞ < 1 – x² < ∞

-∞ < -x² < ∞

-∞ < x < ∞

x ∈ R

Domain = R

Question 2.

Find the value of

(i) tan^{-1}(tan\(\frac {5π}{4}\))

(ii) tan^{-1}(tan(-\(\frac {π}{6}\)))

Solution:

Question 3.

Find the value of

(i) tan(tan^{-1}(\(\frac {7π}{4}\)))

(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(cos^{-1}(\(\frac {1}{2}\)) – sin^{-1}(-\(\frac {1}{2}\)))

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

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

solution: