Question

Sin a + b minus 2 sin a + sin a minus b upon cos a + b - 2 cos a + cos a minus b = tana

Answer 1

HEYA !

$\text{LHS} \\ \\ = \frac{ \sin(a + b) - 2 \sin a + \sin(a - b) }{ \cos(a + b) - 2 \cos a + \cos(a - b) } \\ \\ = \frac{ [ \: \sin(a + b) + \sin(a - b) \: ] - 2 \sin a}{ [ \:\cos(a + b) + \cos(a - b)\: ] - 2 \cos a } \\ \\ = \frac{2 \sin a \: \cos b - 2 \sin a}{2 \cos a \: \cos b - 2 \cos a} \\ \\ = \frac{2 \sin a( \cos b - 1)}{2 \cos a( \cos b - 1)} \\ \\ = \frac{ \sin a}{ \cos b} \\ \\ = \tan a \: \: \: \: \: = \text{RHS} \: \: \: [ \text{ PROVED } ]$
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FORMULA USED :

• sin(x + y) + sin(x - y) = 2 sinx cosy

• cos(x + y) + cos(x - y) = 2cosx cosy

$\boxed{ \mathbf{ \: HOPE \: \: \: IT \: \: \: HELPS \: }}$

Answer 2

Hello friend

Your answer is given in the attachment

I hope it will help you a lot

Thanks

Have a unique day user

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