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Question: If in a triangle ABC, ∠B = 90<sup>0</sup> then tan<sup>2</sup> \(\left( \frac { A } { 2 } \right)\) ...

If in a triangle ABC, ∠B = 900 then tan2 (A2)\left( \frac { A } { 2 } \right) is

A

b−cb+c\frac { b - c } { b + c }

B

b+cb−c\frac { b + c } { b - c }

C

2bcb+c\frac { 2 b c } { b + c }

D

None of these

Answer

b−cb+c\frac { b - c } { b + c }

Explanation

Solution

cosA = cb\frac { c } { b } ⇒ 1−tan⁡2A21+tan⁡2A2=cb\frac { 1 - \tan ^ { 2 } \frac { A } { 2 } } { 1 + \tan ^ { 2 } \frac { A } { 2 } } = \frac { c } { b }

⇒

(by componendo and dividendo rule)

Hence tan2 =b−cb+c\frac { b - c } { b + c }