Question

please give me the answer
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Answer 1

The solution is in the image

hope it helps you :)

Answer 2

${\pmb{\underline{\sf{Required \: Solution....}}}}$

Take LHS !!

$: \implies \frac{ \tan {}^{3}θ }{ \sec {}^{2}θ } + \frac{ \cot {}^{2} θ }{cosec {}^{2} θ} \\ : \implies \frac{ \sin θ \times \sf\red{\cos {}^{2} θ}} { \cos\sf\red{ {}^{3} }θ \times 1 } + \frac{ \cos {}^{3} θ \times \sf\red{\sin{}^{2} } θ }{ \sin \sf\red{ {}^{3} } \times1 } \\ : \implies \frac{ \sin {}^{3}θ }{ \cosθ} + \frac{ \cos {}^{3}θ }{ \sin θ} \\ : \implies \frac{ \sin {}^{4}θ + \cos {}^{4} θ}{ \sinθ \times \cosθ } \\ : \implies \: \frac{ (\sin {}^{2}θ) {}^{2} + ( \cos {}^{2} θ) {}^{2} }{ \sinθ \cosθ }$

We know that ;

$\boxed{a {}^{2} + b {}^{2} + 2ab = (a + b) {}^{2} }$

We can write it as ;

$\large{\boxed{\bf{\green{a {}^{2} + b {}^{2} = (a + b) {}^{2} - 2ab }}}}$

$: \implies \frac{ (\sin {}^{2} θ + \cos{}^{2} θ) {}^{2} - 2 \sin {}^{2} θ \times \cos {}^{2} θ}{ \sinθ \times \cos θ } \\ \implies \frac{1 - 2 \sin {}^{2}θ \cos {}^{2} θ }{ \sin θ \cosθ} \\ \implies \frac{1}{ \sinθ \cosθ } - \frac{2 \sin\sf\red{ {}^{2} }θ \cos \sf\red{{}^{2}}θ }{ \sf\red{ \sinθ \cosθ }} \sf\red{} \\ : \implies \sinθ \times \cosecθ - 2 \sinθ \cosθ \\ = RHS$

Hence Proved ;)

LHS = RHS

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@Mahor2111

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