Question

Heyy guys.....
answer this one Plzz...
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Answer 1

On taking L.H.S. :

${ \tan }^{2} \alpha - { \tan}^{2} \beta \\ \\ = > \frac{ { \sin}^{2} \alpha }{ { \cos }^{2} \alpha } - \frac{ { \sin }^{2} \beta }{ { \cos }^{2} \beta } \\ \\ = > \frac{ { \sin }^{2} \alpha { \cos }^{2} \beta - { \sin }^{2} \beta { \cos}^{2} \alpha }{ { \cos}^{2} \alpha { \cos}^{2} \beta } \\ \\ = > \frac{ { \sin }^{2} \alpha (1 - { \sin}^{2} \beta ) - { \sin}^{2} \beta (1 - { \sin }^{2} \alpha) }{ { \cos}^{2} \alpha { \cos }^{2} \beta } \\ \\ = > \frac{ { \sin}^{2} \alpha - { \sin}^{2} \alpha { \sin }^{2} \beta - { \sin}^{2} \beta + { \sin }^{2} \alpha { \sin }^{2} \beta }{ { \cos }^{2} \alpha { \cos }^{2} \beta } \\ \\ = > \frac{ { \sin}^{2} \alpha - { \sin }^{2} \beta }{ { \cos }^{2} \alpha { \cos}^{2} \beta } = R.H.S.$

HENCE PROVED ✔️✔️

✌️_____ Fóllòw MË _____✌️

Answer 2

Step-by-step explanation:

hope it helps you bye take care