Question

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Answer 1

Answer:-

Given that,

$\sf { \cos}^{2} \theta = \frac{ {a}^{2} - 1}{3}$

$\sf { \tan}^{2} ( \frac{ \theta}{2} ) = { \tan }^{ \frac{2}{3} } \alpha$

$\sf \implies { \tan}^{3} ( \frac{ \theta}{2} ) = \tan\alpha$

$\sf \implies \frac{ { \sin}^{3}( \frac{\theta}{2} )}{{ \cos}^{3}( \frac{\theta}{2} )} = \frac{ \sin \alpha }{ \cos \alpha }$

$\sf \implies \frac{ { \sin}^{3}( \frac{\theta}{2} )}{ \sin\alpha } = \frac{{ \cos}^{3}( \frac{\theta}{2} )}{ \cos \alpha }$

Let us assume that,

$\sf\frac{ { \sin}^{3}( \frac{\theta}{2} )}{ \sin\alpha } = \frac{{ \cos}^{3}( \frac{\theta}{2} )}{ \cos \alpha } = k$

$\sf \implies{ \sin}^{3}( \frac{\theta}{2} ) = k \sin \alpha \: ...(i)$

$\sf \implies{ \cos}^{3}( \frac{\theta}{2} ) = k \cos \alpha \: ...(i)$

$\sf \therefore {k}^{ \frac{2}{3} } { \sin}^{ \frac{2}{3} } \alpha + {k}^{ \frac{2}{3} } { \cos}^{ \frac{2}{3} } \alpha = 1$

Squaring and adding equation (i) and (ii), we get,

$\sf \implies {k}^{2} ( { \sin }^{2} \alpha + { \cos}^{2} \alpha ) = { \sin}^{6} ( \frac{ \theta}{2} ) + { \cos }^{6} ( \frac{ \theta}{2} )$

$\sf \implies {k}^{2} = { \sin}^{6} ( \frac{ \theta}{2} ) + { \cos }^{6} ( \frac{ \theta}{2} )$

$\sf \implies {k}^{2} = {( { \sin}^{2} ( \frac{ \theta}{2} ) + { \cos}^{2}( \frac{ \theta}{2} ))}^{3} - 3 { \sin }^{2}( \frac{ \theta}{2} ){ \cos }^{2}( \frac{ \theta}{2} )({ \sin}^{2} ( \frac{ \theta}{2} ) + { \cos}^{2}( \frac{ \theta}{2} ))$

$\sf \implies {k}^{2} = 1 - \frac{3}{4} ( { \sin}^{2} \theta)$

$\sf \implies {k}^{2} = 1 - \frac{3}{4} ( 1 - {\cos}^{2} \theta)$

$\sf \implies {k}^{2} = 1 - \frac{3}{4} - \frac{3}{4} {\cos}^{2} \theta$

$\sf \implies {k}^{2} = \frac{ {a}^{2} }{4} \implies k = \frac{a}{2}$

$\sf \therefore {k}^{ \frac{2}{3} } { \sin}^{ \frac{2}{3} } \alpha + {k}^{ \frac{2}{3} } { \cos}^{ \frac{2}{3} } \alpha = 1$

$\sf \implies{k}^{ \frac{2}{3} } ( { \sin}^{ \frac{2}{3} } \alpha + { \cos}^{ \frac{2}{3} } \alpha) = 1$

$\sf \implies ( { \sin}^{ \frac{2}{3} } \alpha + { \cos}^{ \frac{2}{3} } \alpha) = \frac{1}{ {k}^{ \frac{2}{3} } }$

$\sf = \large \Big( \frac{1}{ \frac{a}{2} } \Big)^{ \frac{2}{3} }$

$\sf = {( \frac{2}{a})}^{3}$

Hence,

$\boxed{\sf { \sin}^{ \frac{2}{3} } \alpha + { \cos}^{ \frac{2}{3} } \alpha= { \bigg(\frac{2}{a} \bigg)}^{ \frac{2}{3} } }$

Correct Answer:- Option B

This is the required answer.

Answer 2

Answer:

OLA!

Step-by-step explanation:

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