Question

$\sf{ \red{Hey \ Everyone! }}$

Answer this maths question
[Chapter Trigonometry, Class 10]
Prove that:-

$\bf( \dfrac{1}{ {sec}^{2} \alpha - {cos}^{2} \alpha } + \dfrac{1}{ {cosec}^{2} \alpha - {sin}^{2} \alpha})$
$\bf\times {sin}^{2} \alpha \ {cos}^{2} \alpha =$
$\sf{ \purple{\dfrac{1 - {sin}^{2} \alpha \ {cos}^{2} \alpha }{ 2 + {sin}^{2} \alpha \ {cos}^{2} \alpha }}}$
[Note that it has × sign also in LHS, and it is continued. Both LHS and RHS are written in different fonts]

Please, do not spam.
Correct answer with valid explanation will be marked as Brainliest. ​

Answer 1

Formula Applied :

$\sf \bullet\ \; \red{sin^2\theta +cos^2\theta =1}$

$\bullet\ \; \sf \blue{sec\theta =\dfrac{1}{cos\theta}}\\\\\bullet\ \; \sf \pink{csc\theta=\dfrac{1}{sin\theta}}$

$\bullet\ \; \sf \green{a^2-b^2=(a+b)(a-b)}$

Solution :

LHS

$\\ \to \sf \left(\dfrac{1}{sec^2x-cos^2x}+\dfrac{1}{csc^2x-sin^2x}\right)sin^2x.cos^2x$

$\\ \to \sf \left( \dfrac{1}{\frac{1}{cos^2x}-cos^2x}+\dfrac{1}{\frac{1}{sin^2x}-sin^2x}\right)sin^2xcos^2x$

$\\ \to \sf \left(\dfrac{1}{\frac{1-cos^4x}{cos^2x}}+\dfrac{1}{\frac{1-sin^4x}{sin^2x}}\right)sin^2xcos^2x$

$\\ \to \sf \left(\dfrac{cos^2x}{1-cos^4x}+\dfrac{sin^2x}{1-sin^4x}\right)sin^2xcos^2x$

$\\ \to \sf \left(\dfrac{cos^2x(1-sin^4x)+sin^2x(1-cos^4x)}{(1-cos^4x)(1-sin^4x)}\right)sin^2xcos^2x$

$\\ \to \sf \left(\dfrac{cos^2x-cos^2x.sin^4x+sin^2x-sin^2x.cos^4x}{\red{(1+cos^2x)(1-cos^2x)(1+sin^2x)(1-sin^2x)}}\right)sin^2xcos^2x$

$\\ \to \sf \left(\dfrac{(sin^2x+cos^2x)-sin^2x.cos^2x(sin^2x+cos^2x)}{(1+cos^2x)(sin^2x)(1+sin^2x)(cos^2x)}\right)sin^2xcos^2x$

$\\ \to \sf \left(\dfrac{1-sin^2x.cos^2x}{(1+sin^2x)(1+cos^2x)\cancel{(sin^2x.cos^2x)}}\right)\cancel{sin^2x.cos^2x}$

$\\ \to \sf \left(\dfrac{1-sin^2x.cos^2x}{1+(cos^2x+sin^2x)+sin^2x.cos^2x}\right)$

$\\ \leadsto \sf \pink{\dfrac{1-sin^2x.cos^2x}{2+sin^2x.cos^2x}}\ \; \bigstar$

RHS

Note : x = α

Answer 2

To Prove :-

$:\implies\sf \dfrac{1} {sec^2 \alpha -cos^2 \alpha} + \dfrac{1}{ cosec^2\alpha - sin^2\alpha}\times sin^2\alpha \ cos^2 \alpha = \dfrac{1 - {sin}^{2} \alpha \ {cos}^{2} \alpha }{ 2 + {sin}^{2} \alpha \ {cos}^{2} \alpha }$

Proof :-

• taking LHS:-

$\mapsto\sf\ \dfrac{1}{sec^2\alpha-cos^2\alpha}+\dfrac{1}{cosec^2\alpha-sin^2\alpha}\times sin^2\alpha cos^2\alpha\\ \\ \\ \bullet\sf\ \ sec\theta= 1/_{cos\theta}\ \ ;\ \ cosec\theta= 1/_{sin\theta}\\ \\ \\ \\ \mapsto\sf\ \Bigg\{\dfrac{1}{\Big(\frac{1}{cos^2\alpha}\Big)-cos^2\alpha}\Bigg\}+\Bigg\{\dfrac{1}{\Big(\frac{1}{sin^2\alpha}\Big)-sin^2\alpha}\Bigg\}\times \big(sin^2\alpha cos^2\alpha\big)\\ \\ \\ \\ \mapsto\sf\ \Bigg\{\dfrac{1}{\frac{1-cos^4\alpha}{cos^2\alpha}}\Bigg\}+\Bigg\{\dfrac{1}{\frac{1-sin^4\alpha}{sin^2\alpha}}\Bigg\}\times \big( sin^2\alpha cos^2\alpha\big)\\ \\ \\ \\ \mapsto\sf\ \bigg\{\dfrac{cos^2\alpha}{1-cos^4\alpha}\bigg\}+\bigg\{\dfrac{sin^2\alpha}{1-sin^4\alpha}\bigg\}\times \big(sin^2\alpha cos^2\alpha\big)$

$\underline{\bigstar{\sf\ \ a^2-b^2=(a+b)(a-b)}}$

$\\ \\ \mapsto\sf \bigg\{\dfrac{cos^2\alpha}{(1+cos^2\alpha)(1-cos^2\alpha)}\bigg\}+\bigg\{\dfrac{sin^2\alpha}{(1+sin^2\alpha)(1-sin^2\alpha)}\bigg\}\times \big(sin^2\alpha cos^2\alpha\big)\\ \\ \\ \\ \star\sf\ \ sin^2\theta+cos^2\theta=1\\ \\ \\ \\ \mapsto\sf\ \bigg\{\dfrac{cos^2\alpha}{(1+cos^2\alpha)sin^2\alpha}\bigg\}+\bigg\{\dfrac{sin^2\alpha}{(1+sin^2\alpha)cos^2\alpha}\bigg\}\times \big(sin^2\alpha cos^2\alpha\big)\\ \\ \\ \\ \mapsto\sf\ \dfrac{cos^2\alpha\Big\{(1+sin^2\alpha)cos^2\alpha\Big\}+sin^2\alpha\Big\{(1+cos^2\alpha)sin^2\alpha\Big\}}{\cancel{sin^2\alpha cos^2\alpha}(1+cos^2\alpha)(1+sin^2\alpha)}\times \cancel{\big(sin^2\alpha cos^2\alpha\big)}\\ \\ \\ \\ \mapsto\sf\ \dfrac{cos^4\alpha+sin^2\alpha cos^4\alpha+ sin^4\alpha+cos^2\alpha sin^4\alpha}{(1+cos^2\alpha)(1+sin^2\alpha)}$

$\\ \\ \mapsto\sf\ \dfrac{sin^4\alpha+cos^4\alpha+sin^2\alpha cos^4\alpha + sin^4\alpha cos^2\alpha}{(1+cos^2\alpha)(1+sin^2\alpha)}\\ \\ \\ \bullet\sf\ \ (sin^2\alpha+cos^2\alpha)^2-2sin^2\alpha cos^2\alpha= (sin^4\alpha+cos^4\alpha)\\ \\ \\ \bullet\sf\ \ 1-2sin^2\alpha cos^2\alpha= sin^4\alpha+cos^4\alpha \\ \\ \\ \\ \mapsto\sf\ \dfrac{1-2sin^2\alpha cos^2\alpha +sin^2\alpha cos^2\alpha(sin^2\alpha+cos^2\alpha)}{1+sin^2\alpha+cos^2\alpha+sin^2\alpha cos^2\alpha}\\ \\ \\ \\ \mapsto\sf\ \dfrac{1-2sin^2\alpha cos^2\alpha+sin^2\alpha cos^2\alpha(1)}{1+1+sin^2\alpha cos^2\alpha}\ \ \ \ \therefore\bigg[sin^2\theta+cos^2\theta=1\bigg]\\ \\ \\ \\ \mapsto\boxed{\red{\sf \dfrac{1-sin^2\alpha cos^2\alpha}{2+sin^2\alpha cos^2\alpha}}}\ \ \ \ \ \sf\ Hence\ Proved !!$