Question #39a5a

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Question #39a5a

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Answer 1 · 2018-01-12T13:52:16

To prove ${tan}^{2} x - {sin}^{2} x = {tan}^{2} x {sin}^{2} x$ $tan^2x-sin^2x=tan^2xsin^2x$

${tan}^{2} x - {sin}^{2} x$ $tan^2x-sin^2x$
$= \frac{{sin}^{2} x}{{cos}^{2} x} - {sin}^{2} x$ $=(sin^2x)/(cos^2x)-sin^2x$ [Put tangent in terms of sine and cosine]
$= \frac{{sin}^{2} x - {sin}^{2} x {cos}^{2} x}{{cos}^{2} x}$ $=(sin^2x-sin^2xcos^2x)/cos^2x$ [Common denominator]
$= \frac{{sin}^{2} x (1 - {cos}^{2} x)}{{cos}^{2} x}$ $=(sin^2x(1-cos^2x))/cos^2x$ [Factorise out the ${sin}^{2} x$ $sin^2x$ ]
$= \frac{{sin}^{2} x \cdot {sin}^{2} x}{{cos}^{2} x}$ $=(sin^2x*sin^2x)/(cos^2x)$ [Since ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$ , $1 - {cos}^{2} x = {sin}^{2} x$ $1-cos^2x=sin^2x$ ]
$= {tan}^{2} x {sin}^{2} x$ $=tan^2xsin^2x$ [Since $\frac{{sin}^{2} x}{{cos}^{2}} = {tan}^{2} x$ $sin^2x/(cos^2)=tan^2x$ ]

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Question #39a5a

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