2tan^-1 (-3) is equal to a -cos^-1 (-4/5) b -pi+cos^-1 (4/5) c -pi/2 +tan^-1(-4/3) d cot^-1(4/3) - Maths - Inverse Trigonometric Functions

Question

2tan^-1 (-3) is equal to a -cos^-1 (-4/5) b -pi+cos^-1 (4/5) c
-pi/2 +tan^-1(-4/3) d cot^-1(4/3)

Ajanta Trivedi · Accepted Answer

let $\tan^{- 1} (- 3) = α \Rightarrow \tan α = - 3$ and

$- \frac{π}{2} < α < 0 \Rightarrow - π < 2 α < 0$ ..........(1)

since $\cos (- 2 α) = \cos 2 α$

$= \frac{1 - \tan^{2} α}{1 + \tan^{2} α} = \frac{1 - (- 3)^{2}}{1 + (- 3)^{2}} = \frac{- 4}{5} \cos (2 α) = \cos (- 2 α) = - \frac{4}{5}$ ....................................(2)

therefore

$- 2 α = \cos^{- 1} (- \frac{4}{5}) 2 \tan^{- 1} (- 3) = - \cos^{- 1} (- \frac{4}{5})$ .................(3)

again $- π < 2 α < 0 \Rightarrow 0 < π + 2 α < π$

therefore

$\cos (π + 2 α) = - \cos 2 α = \frac{4}{5} f r o m e q (2) π + 2 α = \cos^{- 1} (\frac{4}{5}) 2 α = - π + \cos^{- 1} (\frac{4}{5}) 2 \tan^{- 1} (- 3) = - π + \cos^{- 1} (\frac{4}{5})$ ...............(4)

now again $- π < 2 α < 0 \Rightarrow - \frac{π}{2} < \frac{π}{2} + 2 α < \frac{π}{2}$

therefore

$\tan (\frac{π}{2} + 2 α) = - \cot 2 α = - \frac{1}{\tan 2 α} = - \frac{1 - \tan^{2} α}{2 \tan α} = - \frac{1 - {(- 3)}^{2}}{2 * (- 3)}$

$= - \frac{1 - 9}{- 6} = \frac{- 4}{3} \Rightarrow \frac{π}{2} + 2 α = \tan^{- 1} (- \frac{4}{3}) 2 α = - \frac{π}{2} + \tan^{- 1} (- \frac{4}{3})$

$2 \tan^{- 1} (- 3) = - \frac{π}{2} + \tan^{- 1} (- \frac{4}{3})$ .............(5)

thus from (3), (4) and (5), option (a), (b) and (c) are correct.

hope this helps you.

cheers!!

2tan^-1 (-3) is equal to a. -cos^-1 (-4/5) b.-pi+cos^-1 (4/5) c.-pi/2 +tan^-1(-4/3) d.cot^-1(4/3)