Question

Solve the problem!! ​

Answer 1

${ (\sin \: theta + cosec \: theta)}^{2} + {(cos \: theta + sec \: theta})^{2} \\ = > {sin}^{2}theta + {cosec}^{2} theta + 2sin \: theta \times \: cosec \: theta + {cos}^{2} theta + {sec}^{2} theta + 2cos \: theta \: \times sec \: theta \\ = > {sin}^{2} theta \: + {cos}^{2} theta \: + {cosec}^{2} theta + {sec}^{2} \: theta + 2sin \: theta \times cosec \: theta + 2cos \: theta \times sec \: theta \\ = > 1 + 1 + {cot}^{2} theta + 1 + {tan}^{2} theta + 2sin \: theta \times \frac{1}{sin \: theta} + 2cos \: theta \times \frac{1}{cos \: theta} \\ = > 1 + 1 + {cot}^{2} theta + 1 + {tan}^{2} theta + 2 + 2 \\ = > 7 + {tan}^{2} theta + {cot}^{2} theta \: = rhs$

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Answer 2

$\underline{ \underline \mathfrak{answer - }}$

We have,

LHS =

$( { \sin \theta + \cosec \theta ) }^{2} + ( { \cos\theta + \sec \theta) }^{2} \\ = ( { \sin }^{2} \theta + { \cosec }^{2} \theta + 2 \sin \theta \cosec \theta) + ( { \cos }^{2} \theta + { \sec}^{2} \theta + 2 \cos \theta \sec \theta \\ = ({ \sin }^{2} \theta + { \cosec }^{2} \theta + 2) + ( { \cos }^{2} \theta + { \sec}^{2} \theta + 2) \\ = ({ \sin }^{2} \theta + { \cos }^{2} \theta) + 4 + ( { \cosec }^{2} \theta + { \sec}^{2} \theta) \\ = 1 + 4 + (1 + { \cot }^{2} \theta) + (1 + { \tan}^{2} \theta) \\ = (7 + { \tan }^{2} \theta + { \cot }^{2} \theta) = rhs$

Hence, LHS = RHS

$\huge \underline{ \underline \mathfrak{proved}}$