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Question: tan<sup>–1</sup>![](https://cdn.pureessence.tech/canvas_28.png?top_left_x=1365&top_left_y=1087&width...

tan–1+ tan (π412cos1(3ab))\left( \frac { \pi } { 4 } - \frac { 1 } { 2 } \cos ^ { - 1 } \left( \frac { 3 \mathrm { a } } { \mathrm { b } } \right) \right) is equal to –

A

B

C

D

32ab\frac { 3 } { 2 } \frac { \mathrm { a } } { \mathrm { b } }

Answer

Explanation

Solution

Let cos–1 (3ab)\left( \frac { 3 a } { b } \right) = q Ž cos q = (3ab)\left( \frac { 3 a } { b } \right)

Now tan (π4+θ2)\left( \frac { \pi } { 4 } + \frac { \theta } { 2 } \right) + tan (π4θ2)\left( \frac { \pi } { 4 } - \frac { \theta } { 2 } \right)

= 1+tanθ21tanθ2\frac { 1 + \tan \frac { \theta } { 2 } } { 1 - \tan \frac { \theta } { 2 } } + 1tanθ21+tanθ2\frac { 1 - \tan \frac { \theta } { 2 } } { 1 + \tan \frac { \theta } { 2 } }

= 2(1+tan2(θ2))1tan2(θ2)\frac { 2 \left( 1 + \tan ^ { 2 } \left( \frac { \theta } { 2 } \right) \right) } { 1 - \tan ^ { 2 } \left( \frac { \theta } { 2 } \right) } = 2cosθ\frac { 2 } { \cos \theta }

= = 23(ba)\frac { 2 } { 3 } \left( \frac { b } { a } \right) .