Question

Helo help me prove question

Answer 1

hello..



please refer the above attachment...




thank you...
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Answer 2

hii. ....friend

the answer is here,

I take lambda as x.

Sin2A =x Sin2B.

=> Sin2A/Sin2B= x/1.

By componendo & dividendo.


$= > \: \frac{ \sin(2a) + \sin(2b) }{ \sin(2a) - \sin(2b) } = \frac{x + 1}{x - 1}$

By using formulas,

$= > \: \sin(c ) + \sin(d) = 2 \sin( \frac{c + d}{2} ) . \cos( \frac{c - d}{2} )$
And

$= > \: \sin(c) - \sin(d) = 2 \sin( \frac{c - d}{2}). \cos( \frac{c + d}{2})$
We get ,

$= > \: \frac{2 \sin( \frac{2a + 2b}{2}) \cos( \frac{2a - 2b}{2} ) }{2 \sin( \frac{2a - 2b}{2} ) \cos( \frac{2a + 2b}{2} ) } = \frac{x + 1}{x - 1}$

$= > \: \frac{ \sin(a + b) \cos(a - b) }{ \cos(a + b) \sin(a - b) } = \frac{x + 1}{x - 1}$

$= > \: \tan(a + b) . \cot(a - b) = \frac{x + 1}{x - 1}$

$= > \: \frac{ \tan(a + b) }{ \tan(a - b) } = \frac{x + 1}{x - 1}$


:-)Hope it helps u.