Question

Prove the following:
1.
$tan ^{2} \alpha - \sin ^{2} \alpha = \tan ^{2} \alpha \times \sin ^{2} \alpha$
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Answer 1

Step-by-step explanation:

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Answer 2

Solution :

$\tt \implies tan ^{2} \alpha - \sin ^{2} \alpha = \tan ^{2} \alpha \times \sin ^{2} \alpha \\ \\ \tt \implies \frac{ { \sin }^{2} \alpha }{ { \cos}^{2} \alpha } - \sin ^{2} \alpha = \tan ^{2} \alpha \times \sin ^{2} \alpha \\ \\ \tt \implies \frac{ { \sin }^{2} \alpha - \sin ^{2} \alpha. { \cos}^{2} \alpha }{ { \cos}^{2} \alpha } = \tan ^{2} \alpha \times \sin ^{2} \alpha \\ \\\tt \implies \frac{ { \sin }^{2} \alpha (1 - { \cos}^{2} \alpha) }{ { \cos}^{2} \alpha } = \tan ^{2} \alpha \times \sin ^{2} \alpha \\ \\ \tt \implies \tan ^{2} \alpha \times \sin ^{2} \alpha = \tan ^{2} \alpha \times \sin ^{2} \alpha$

L.H.S = R.H.S

Hence Proved

Identities Used :

$\implies \tt \dfrac{ { \sin}^{2} \theta}{ { \cos}^{2}\theta } = { \tan}^{2} \theta \\ \\ \implies \tt 1 - { \sin}^{2} \theta = { \cos}^{2} \theta$