Question

please solve the above question with explanation

Answer 1

The required value is 2..
hope it helps you...

Answer 2

Hope u like my process
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$= > \sin( \theta ) + \cos( \theta ) = p \\ \\ = > \sec( \theta ) + \csc( \theta ) = q \\ \\ or. \: \: \frac{1}{ \cos( \theta ) } + \frac{1}{ \sin( \theta ) } = q \\ \\ or . \: \: \frac{ \sin( \theta ) + \cos( \theta ) }{ \sin( \theta ) \cos( \theta ) } = q \\ \\ or. \: \: \frac{p}{ \sin( \theta ) \cos( \theta ) } = q \\ \\ or. \: \: \frac{q}{p} = \frac{1}{ \sin( \theta ) \cos( \theta ) }$
Now,

$= > \frac{q}{p} ( {p}^{2} - 1) \\ \\ = \frac{1}{ \sin( \theta ) \cos( \theta ) } ( {( \sin( \theta ) + \cos( \theta )) }^{2} - 1) \\ \\ = \frac{1}{ \sin( \theta ) \cos( \theta ) } ( (\sin ^{2} ( \theta ) + \cos ^{2} ( \theta ) - 2 \sin( \theta ) \cos( \theta ) ) - ( \sin ^{2} ( \theta ) + \cos ^{2} ( \theta ) )) \\ \\ = \frac{2 \sin( \theta ) \cos( \theta ) }{ \sin( \theta ) \cos( \theta ) } = 2$
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Hope this is ur required answer

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