Question

solve tan^theta- sin^theta= tan^theta sin^theta

Answer 1

you can see your answer

Answer 2

$= > {tan}^{2} \alpha - {sin}^{2} \alpha = {tan}^{2} \alpha . {sin}^{2} \alpha \\ \\ = > \frac{ {sin}^{2} \alpha }{ {cos}^{2} \alpha } - {sin}^{2} \alpha = {tan}^{2} \alpha . {sin}^{2} \alpha \\ \\ = > \frac{ {sin}^{2} \alpha (1 - {cos}^{2} \alpha ) }{ {cos}^{2} \alpha } = {tan}^{2} \alpha . {sin}^{2} \alpha$
______________________________

TRIGONOMETRY IDENTIFY :-

$= > tan \alpha = \frac{sin \alpha }{cos \alpha } \\ \\ = > 1 - {cos}^{2} \alpha = { sin}^{2} \alpha$
________________________________

THEN,

$= > \frac{ {sin}^{2} \alpha (1 - {cos}^{2} \alpha ) }{ {cos}^{2} \alpha } = {tan}^{2} \alpha . {sin}^{2} \alpha \\ \\ = > {tan}^{2} \alpha . {sin}^{2} \alpha = {tan}^{2} \alpha . {sin}^{2} \alpha$
____________________"[PROVED]"

=======================================

_-_-_-_☆$BE \: \: \: \: BRAINLY$☆_-_-_-_