Question

plz solve it
Urgently

Answer 1

Solution =taking LHS,
$\frac{ \tan(a) + \tan(b) }{ \tan(a) - \tan(b) } \\ \\ = \frac{ \sin(a) }{ \cos(a) } + \frac{ \sin(b) }{ \cos(b) } \div \frac{ \sin(a) }{ \cos(a) } - \frac{ \sin(b) }{ \cos(b) } \\ = \frac{ \sin(a) \cos(b) + \cos(a) \sin(b) }{ \cos(a) \cos(b) } \div \frac{ \sin(a) \cos(b) - \cos(a) \sin(b) }{ \cos(a) \cos(b) } \\ = \frac{ \sin(a + b) }{ \sin(a - b) }$
RHS

Here proved✌✌✌

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Answer 2

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$\frac{tan\: A+ tan\: B}{tan\: A-tan\: B}=\frac{Sin\: (A+B)}{Sin\: (A-B)}$

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$\large{\blue{\underline{\blue{LHS=}}}}$

$\frac{tan\: A+ tan\: B}{tan\: A-tan\: B}$

Since $tan\: \theta = \frac{Sin\: \theta}{Cos\: \theta}$

= $\frac{\frac{Sin\: A}{Cos\: A}+ \frac{Sin\: B}{Cos\: B}}{\frac{Sin\: A}{Cos\: A}- \frac{Sin\: B}{Cos\: B}}$

= $\frac{Sin\: A.Cos\: B+ Cos\: A.Sin\: B}{Sin\: A.Cos\: B- Cos\: A.Sin\: B}$

We have,
$Sin\: (A+B) = Sin\: A.Cos\: B+ Cos\: A.Sin\: B$ &
$Sin\: (A-B) = Sin\: A.Cos\: B - Cos\: A.Sin\: B$

=> $\frac{Sin\: (A+B)}{Sin\: (A-B)}$

$\large{\blue{\underline{\blue{=RHS}}}}$

✓✓✓

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