Question

sec^2A×cosec^2=tan^2A+cot^2A+2

Answer 1

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Answer 2

The answer is given below :

Now, L.H.S.

$= {sec}^{2} \alpha \times {cosec}^{2} \alpha \\ \\ = \frac{1}{ {cos}^{2} \alpha } \times \frac{1}{ {sin}^{2} \alpha } \\ \\ = \frac{1}{ {cos}^{2} \alpha \: \: {sin}^{2} \alpha } \\ \\ = \frac{ {( {sin}^{2} \alpha + {cos}^{2} \alpha ) }^{2} }{ {cos}^{2} \alpha \: \: {sin}^{2} \alpha } \\ \\ = \frac{ {sin}^{4} \alpha + {cos}^{4} \alpha + 2 {sin}^{2} \alpha \: \: {cos}^{2} \alpha }{ {sin}^{2} \alpha \: \: {cos}^{2} \alpha } \\ \\ = \frac{ {sin}^{2} \alpha }{ {cos}^{2} \alpha } + \frac{ {cos}^{2} \alpha }{ {sin}^{2} \alpha } + 2 \\ \\ = {tan}^{2} \alpha + {cot}^{2} \alpha + 2$

= R.H.S. [Proved]

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