How to prove this identity? ${sin}^{2} x + {tan}^{2} x \cdot {sin}^{2} x = {tan}^{2} x$ $sin^2x+tan^2x * sin^2x= tan^2x$

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How to prove this identity? ${sin}^{2} x + {tan}^{2} x \cdot {sin}^{2} x = {tan}^{2} x$ $sin^2x+tan^2x * sin^2x= tan^2x$

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Answer 1 · 2018-03-18T12:59:24

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Explanation:

Use our trig identities...

${sin}^{2} x + {cos}^{2} x = 1$ $sin^2 x + cos^2 x =1$

$\Rightarrow \frac{{sin}^{2} x}{{cos}^{2} x} + \frac{{cos}^{2} x}{{cos}^{2} x} = \frac{1}{{cos}^{2} x}$ $=> sin^2 x / cos^2 x + cos^2 x / cos^2 x = 1 / cos^2 x$

$\Rightarrow {tan}^{2} x + 1 = \frac{1}{{cos}^{2} x}$ $=> tan^2 x + 1 = 1/cos^2 x$

Factor the left side of your problem...

$\Rightarrow {sin}^{2} x (1 + {tan}^{2} x)$ $=> sin^2 x ( 1 + tan^2 x )$

$\Rightarrow {sin}^{2} x (\frac{1}{{cos}^{2} x}) = \frac{{sin}^{2} x}{{cos}^{2} x}$ $=> sin^2 x ( 1/cos^2 x ) = sin^2 x / cos^2 x$

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

Answer 2 · 2018-03-18T13:04:31

Given,

${sin}^{2} x + {tan}^{2} x {sin}^{2} x$ $sin^2 x +tan^2x sin^2x$

$= {sin}^{2} x (1 + {tan}^{2} x)$ $=sin^2 x(1+tan^2 x)$

$= {sin}^{2} x {sec}^{2} x$ $=sin ^2 x sec ^2x$ (as, ${sec}^{2} x - {tan}^{2} x = 1)$ $sec^2x - tan^2 x=1)$

$= {sin}^{2} x (\frac{1}{{cos}^{2} x})$ $= sin ^2x (1/(cos ^2x))$

$= {tan}^{2} x$ $=tan^2 x$

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How to prove this identity? ${sin}^{2} x + {tan}^{2} x \cdot {sin}^{2} x = {tan}^{2} x$ $sin^2x+tan^2x * sin^2x= tan^2x$

2 Answers

Explanation:

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How to prove this identity? sin2x+tan2x⋅sin2x=tan2xsin^2x+tan^2x * sin^2x= tan^2x

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How to prove this identity? ${sin}^{2} x + {tan}^{2} x \cdot {sin}^{2} x = {tan}^{2} x$ $sin^2x+tan^2x * sin^2x= tan^2x$