Question

Prove that :

$\bigstar \rm{ \: \tan {}^{2} \theta - \sin {}^{2} \theta = \sin {}^{4} \theta \sec {}^{2} \theta }$
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Answer 1

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Answer 2

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\: {tan}^{2}\theta - {sin}^{2}\theta = {sin}^{4}\theta {sec}^{2}\theta$

$\large\underline{\sf{Solution-}}$

Consider, LHS

$\rm :\longmapsto\: {tan}^{2}\theta - {sin}^{2}\theta$

We know

$\boxed{ \bf{ \: tanx = \frac{sinx}{cosx}}}$

So, using this identity, we get

$\rm \:  =  \:\dfrac{ {sin}^{2} \theta }{ {cos}^{2} \theta } - {sin}^{2}\theta$

$\rm \:  =  \:\dfrac{ {sin}^{2} \theta - {sin}^{2}\theta {cos}^{2} \theta}{ {cos}^{2} \theta}$

$\rm \:  =  \:\dfrac{ {sin}^{2} \theta(1 - {cos}^{2} \theta)}{ {cos}^{2} \theta}$

We know,

$\boxed{ \bf{ \: {sin}^{2}x + {cos}^{2}x = 1}}$

So, using this identity, we get

$\rm \:  =  \:\dfrac{ {sin}^{2} \theta \times {sin}^{2} \theta }{ {cos}^{2} \theta }$

$\rm \:  =  \:\dfrac{ {sin}^{4} \theta }{ {cos}^{2} \theta }$

We know,

$\boxed{ \bf{ \: \frac{1}{cos\theta } = sec\theta }}$

So, using this, we get

$\rm \:  =  \: {sin}^{4}\theta {sec}^{2}\theta$

Hence,

$\rm :\longmapsto\:\boxed{ \bf{ \: {tan}^{2}\theta - {sin}^{2}\theta = {sin}^{4}\theta {sec}^{2}\theta}}$

Alternative Method :-

Consider, LHS

$\rm :\longmapsto\: {tan}^{2}\theta - {sin}^{2}\theta$

We know,

$\boxed{ \bf{ \: tanx = \frac{sinx}{cosx}}}$

So, using this,

$\rm \:  =  \:\dfrac{ {sin}^{2} \theta }{ {cos}^{2} \theta } - {sin}^{2}\theta$

$\rm \:  =  \bigg[\:\dfrac{1}{ {cos}^{2} \theta } - 1\bigg] {sin}^{2}\theta$

$\rm \:  =  \bigg[\: {sec}^{2} \theta - 1\bigg] {sin}^{2}\theta$

We know,

$\boxed{ \bf{ \: {sec}^{2}x - {tan}^{2}x = 1}}$

So, using this, we get

$\rm \:  =  \bigg[\: {tan}^{2} \theta\bigg] {sin}^{2}\theta$

$\rm \:  =  \:\dfrac{ {sin}^{2} \theta }{ {cos}^{2} \theta } \times {sin}^{2} \theta$

$\rm \:  =  \:\dfrac{ {sin}^{4} \theta }{ {cos}^{2} \theta }$

$\rm \:  =  \: {sin}^{4}\theta {sec}^{2}\theta$

Hence,

$\rm :\longmapsto\:\boxed{ \bf{ \: {tan}^{2}\theta - {sin}^{2}\theta = {sin}^{4}\theta {sec}^{2}\theta}}$