IIT JEE 1983 Maths - True or False Q9

Relevant wiki: Half Angle Tangent Substitution

$\begin{aligned} \tan A & = \frac {1-\cos B}{\sin B} & \small \color{#3D99F6} \text{Let } t= \tan \frac B2 \\ & = \frac {1-\frac {1-t^2}{1+t^2}}{\frac {2t}{1+t^2}} \\ & = \frac {1+t^2-1+t^2}{2t} \\ & = t = \tan \frac B2 \\ \implies A & = \frac B2 \\ \implies \tan 2A & = \tan B \end{aligned}$

Therefore, the answer is $\boxed{\text{True}}$ .

If we get tan A = cosecB - cot B

Then we get cot A = cosec B + cotB

Subtracting 1st from second we get cot A - tan A = 2/tanB

Hence $1-tan^2A$ / tan A = 2/ tan B

Hence tan B = 2tanA/ ( $1-tan^2A$ )

Hence tan B = tan 2A

Given statement correct.