Question

Sin 48 • sec 4 2 + cos 48 cosec 42
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Answer 1

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Answer 2

SOLUTION

TO DETERMINE

$\sf{} \sin {48}^{ \circ} \sec {42}^{ \circ} + \cos {48}^{ \circ} \cosec {42}^{ \circ}$

FORMULA TO BE IMPLEMENTED

$\sf{}1. \: \: \sec ({90}^{ \circ} - \theta ) = \cosec \theta$

$\sf{}2. \: \: \cosec ({90}^{ \circ} - \theta ) = \sec \theta$

EVALUATION

$\sf{} \sin {48}^{ \circ} \sec {42}^{ \circ} + \cos {48}^{ \circ} \cosec {42}^{ \circ}$

$= \sf{} \sin {48}^{ \circ} \sec ({90}^{ \circ} - {48}^{ \circ} ) + \cos {48}^{ \circ} \cosec ({90}^{ \circ} - {48}^{ \circ} )$

$= \sf{} \sin {48}^{ \circ} \cosec {48}^{ \circ} + \cos {48}^{ \circ} \sec {48}^{ \circ}$

$\displaystyle= \sf{} \sin {48}^{ \circ} \times \frac{1}{\sin {48}^{ \circ}} + \cos {48}^{ \circ} \times \frac{1}{\cos {48}^{ \circ} }$

$\sf{} = 1 + 1$

$= \sf{}2$

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