Question

A, B, c are interior angle of triangle ABC. Then cosec (A+B/2) =?

Answer 1

$\huge \mathfrak \green{ \bigstar{ \star{heya}}}$

$\huge \mathfrak \red { \bigstar{ \underline{ answer}}}$

A+B+C=180°

*{triangle sum property}.

$\frac{A}{2} + \ \frac{B + c}{2} = 90 { \degree} \\ \\ \frac{B + c}{2} = 90 - \frac{A}{2} \\ \\ \sin( \frac{B + c}{2} ) = \sin(90 - \frac{A}{2} )$

$\sin( \frac{B + c}{2} ) = \cos( \frac{A}{2} )$

[sin (90°-Q)=cos Q]

Answer 2

Answer:

Heya mate

Your answer

[sin (90°-Q)=cos Q]

Hope it helps