Question

Solve it plz...
$\frac{\sin(a) - 2 \sin ^{3} (a) }{2 \cos ^{3} (a) - \cos(a) } = \tan(a)$
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Answer 1

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$~~~~~~~~~~~$ $\bigstar\underline\bold\color{red}{TO~ PROVE: }$

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$\longrightarrow{}$$\dfrac{sin \: A - 2sin^{3} \: A}{2cos^{3} A-cos \: A}=tan \: A$

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$~~~~~~~~~~~$ $\bigstar\underline\bold\color{red}{EXPLANATION:}$

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$\longrightarrow{}$$\dfrac{sin \: A - 2sin^{3} \: A}{2cos^{3} A-cos \: A}=tan \: A$

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$\longrightarrow{}$$\dfrac{sin\: A(1 - 2sin^{2} \: A)}{cos\: A(2cos^{2} A-1)}=tan \: A$

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$\longrightarrow{}$$\boxed{ \large{ \bold{ \frac{sin\: A}{cos\: A} = tan\: A }}}$

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$\longrightarrow{}$$\boxed{ \large{ \bold{ 1 - {sin}^{2} \: A = {cos}^{2}\: A }}}$

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$\longrightarrow{}$$\sf{tan\: A \left(\dfrac{1 - 2sin^{2} \: A}{2(1 -{sin}^{2} \: A )-1}\right)=tan \: A}$

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$\longrightarrow{}$$\sf{tan\: A \left(\dfrac{1 - 2sin^{2} \: A}{2-2{sin}^{2} \: A -1} \right)=tan \: A}$

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$\longrightarrow{}$$\sf{tan\: A \left(\dfrac{1 - 2sin^{2} \: A}{1-2{sin}^{2} \: A} \right)=tan \: A}$

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Thankyou :)