Question

$\frac{1 + 2 {sin}^2a }{1 + 3 { \tan }^{2} a} = { \cos }^{2} a$
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Answer 1

Step-by-step explanation:

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Answer 2

SOLUTION:-

Take L.H.S

$= > \frac{1 + 2 {sin}^{2}A }{1 + 3 {tan}^{2} A } \\ \\ = > \frac{1 + 2 {sin}^{2} A }{1 + 3( \frac{ {sin}^{2} A }{ {cos}^{2} A }) } \\ \\ = > \frac{1 + 2 {sin}^{2} A}{ \frac{ {cos}^{2} A + 3 {sin}^{2} A }{ {cos}^{2} A} } \\ \\ = > \frac{(1 + 2 {sin}^{2} A) {cos}^{2} A }{ {cos}^{2} A + {sin}^{2} A + 2 {sin}^{2} A } \\ \\ = > \frac{(1 + 2 {sin}^{2} A) {cos}^{2} A}{(1 +2 {sin}^{2} A) } \: \: \: \: \: \: \: \: [ {cos}^{2} \theta + {sin}^{2} \theta = 1] \\ \\ = > {cos}^{2} A \: \: \: \: \: \: \: \: \: \: [R.H.S.]$

Hence,

Proved.

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