Question

can you please solve the question given below​

Answer 1

Answer:

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Answer 2

Answer:

$\sin( \alpha ) = \frac{ {p}^{2} - 1}{ {p}^{2} + 1 }$

RHS

$\frac{{( \sec\alpha + \tan \alpha )}^{2} - 1 } {( \sec\alpha + \tan \alpha) ^{2} + 1 } \\ let \: (\sec \alpha + \tan\alpha ) \: be \: x \\ = \frac{ {x}^{2} + \tan^{2} \alpha - \sec^{2} \alpha }{ {x}^{2} + \sec ^{2} \alpha - \tan ^{2} \alpha } \\ = \frac{ {x}^{2} + ( \tan\alpha + \sec\alpha )( \tan\alpha - \sec \alpha ) }{ {x}^{2} + x( \sec\alpha - \tan\alpha ) } \\ =let \: tan \alpha - sec \alpha \: be \: y \\ = \frac{ {x}^{2} + xy}{ {x}^{2} - xy } \\ = \frac{x(x + y)}{ {x}(x - y) } \\ = \frac{ 2\tan \alpha }{2 \sec\alpha } \\ = \ \frac{ \sin \alpha }{ \cos\alpha } \times \cos \alpha \\ = \sin \alpha$

Therefore LHS = RHS ,

Hence Verified