Prove the identity Tan^2x-sin^2x is same as (tan^2x)(sin^2x)?

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Prove the identity Tan^2x-sin^2x is same as (tan^2x)(sin^2x)?

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Answer 1 · 2018-04-06T06:54:15

${tan}^{2} x - {sin}^{2} x$ $Tan^2x-sin^2x$

$\Rightarrow \frac{{sin}^{2} x}{{cos}^{2} x} - {sin}^{2} x$ $=> sin^2x/cos^2x-sin^2x$ $w w w w w w w w w$ $color(white)(wwwwwwwww$ $[as tan x = \frac{sin x}{cos x}]$ $["as " tanx = sinx/cosx]$

$\Rightarrow \frac{{sin}^{2} x - {sin}^{2} x {cos}^{2} x}{{cos}^{2} x}$ $=> (sin^2x - sin^2xcos^2x)/cos^2x$

$\Rightarrow \frac{{sin}^{2} x (1 - {cos}^{2} x)}{{cos}^{2} x}$ $=> (sin^2x(1 - cos^2x))/cos^2x$ $w w w w w w$ $color(white)(wwwwww$ $[as {sin}^{2} x + {cos}^{2} x = 1]$ $["as " sin^2x + cos^2x =1]$

$\Rightarrow \frac{{sin}^{2} x ({sin}^{2} x)}{{cos}^{2} x}$ $=> (sin^2x(sin^2x))/cos^2x$

$\Rightarrow {tan}^{2} x {sin}^{2} x$ $=>tan^2xsin^2x$

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Prove the identity Tan^2x-sin^2x is same as (tan^2x)(sin^2x)?

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