Solveeit Logo
Login

Question

Mathematics Question on Inverse Trigonometric Functions

3 tan⁡−1a3\,\tan^{-1}a is equal to

A

tan⁡3a+a31+3a2\tan\frac{3a+a^3}{1+3a^2}

B

tan⁡−13a−a31+3a2\tan^{-1}\frac{3a-a^3}{1+3a^2}

C

tan⁡−13a+a31−3a2\tan^{-1}\frac{3a+a^3}{1-3a^2}

D

tan⁡−13a−a31−3a2\tan^{-1}\frac{3a-a^3}{1-3a^2}

Answer

tan⁡−13a−a31−3a2\tan^{-1}\frac{3a-a^3}{1-3a^2}

Explanation

Solution

We know that tan 3A=3 tan A−tan3A1−3 tan2A tan \,3A =\frac{ 3\,tan\,A-tan^{3}A}{1-3\,tan^{2}A} ∴3 A=tan−13 tan A−tan3A1−3tan2A\therefore 3\,A = tan^{-1} \frac{3\,tan\,A-tan^{3}A}{1-3tan^{2}A} Put tan A=atan\, A = a i.e., A=tan−1aA = tan^{-1}a ∴3tan−1a=tan−13a−a31−3a2\therefore 3tan^{-1} a = tan^{-1} \frac{3a-a^{3}}{1-3a^{2}}