Question

please explain the answer correctly correctly correctly.........................

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Answer 1

Answer:

Hope you will understand, once you see the solution.

Answer 2

$\huge\mathfrak\red{Answer}$

$a( \tan \alpha + \cot \alpha ) = 1 \\ \\ a( \tan \alpha + \frac{1}{ \tan \alpha } ) = 1 \\ \\ a( \frac{ {tan}^{2} \alpha + 1 }{ \tan \alpha } ) = 1 \\ \\ a( \frac{ {sec}^{2} \alpha }{ \tan \alpha } ) = 1 \\ \\ a( \frac{ {sec}^{2} \alpha . \cos \alpha }{ \sin \alpha } ) = 1 \\ \\ a( \frac{ \sec \alpha }{ \sin \alpha } ) = 1 \\ \\ a( \frac{1}{ \sin \alpha . \cos \alpha } ) = 1 \\ \\ \\ a = \sin \alpha \cos \alpha \\ \\ 2a = 2 \sin \alpha . \cos \alpha......(1) \\ \\ b = \sin \alpha + \cos \alpha \\ \\ {b}^{2} = {(sin \alpha + \cos \alpha ) }^{2} \\ \\ {b}^{2} = {sin}^{2} \alpha + {cos}^{2} \alpha + 2 \sin \alpha \cos \alpha \\ \\ {b}^{2} = 1 + 2 \sin \alpha \cos \alpha \\ \\ {b}^{2} - 1 = 2 \sin \alpha \cos \alpha ....(2) \\ \\ from \: (1) \: and \: (2) \: we \: get \\ \\ 2a = {b}^{2} - 1$