Question

tan2 x – sin2x = tan2 x sin2 x​

Answer 1

We have to prove that (tan x)^2 - (sin x)^2 = (tan x)^2 * (sin x)^2

Start from the left hand side:

(tan x)^2 - (sin x)^2

use tan x = sin x / cos x

(sin x)^2/(cos x)^2 - (sin x)^2

=> [(sin x)^2 - (sin x)^2*(cos x)^2]/(cos x)^2

=> (sin x)^2[1 - (cos x)^2]/(cos x)^2

=> (sin x)^2 * [(sin x)^2 / (cos x)^2]

=> (tan x)^2 * (sin x)^2

which is the right hand side.

This proves that (tan x)^2 - (sin x)^2 = (tan x)^2 * (sin x)^2

HOPE THIS HELPS PLZ MARK AS BRAINLIEST

Answer 2

RHS

$= {tan}^{2} x \: {sin}^{2} x \\ = \frac{ {sin}^{2}x }{ {cos}^{2}x } . {sin}^{2} x \\ = \frac{ {sin}^{4}x }{ {cos}^{2} x}$

LHS

$= {tan}^{2} x - {sin}^{2} x \\ = \frac{ {sin}^{2} x}{ {cos}^{2} x} - {sin}^{2} x \\ = \frac{ {sin}^{2} x - {sin}^{2}x \: {cos}^{2} x}{ {cos}^{2}x } \\ = \frac{ {sin}^{2} x(1 - {cos}^{2}x) }{ {cos}^{2}x } \\ = \frac{ {sin}^{2}x \: {sin}^{2}x }{ {cos}^{2} x} \: \: \: \: \: \: \: ... \: ( {sin}^{2} x + {cos}^{2} x = 1) \\ = \frac{ {sin}^{4} x}{ {cos}^{2}x }$

= RHS

since, LHS = RHS

hence,

${tan}^{2} x - {sin}^{2} x = {tan}^{2} x \: {sin}^{2} x$