Question

sin / sin (90 - A) + cos A /( 90 - A ) = sec A cosec A​

Answer 1

Step-by-step explanation:

=> sinA/sin(90-A)+cosA/cos(90-A)=secA cosecA

=> sinA/cosA+cosA/sinA

=> cos^2A+sin^2A/cosA sinA

=> 1/cosA sinA

=> 1/cosA 1/sinA

=> secA cosecA..

Therefore: LHS=RHS....

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Answer 2

Question :

To Prove : sinA/sin(90-A) + cosA/cos(90-A) = secA cosecA

Explanation :

Take L.H.S

$\implies {\sf{\dfrac{\sin A}{\sin (90 \: - \: A)} \: + \: \dfrac{\cos A}{\cos (90 \: - \: A)}}} \\ \\ \footnotesize {\tt{\dag \: \: \: \: \: \sin(90 \: - \: A) \: = \: \cos A \: \: \: \: \: \: and \: \: \: \: \: \: \: \cos(90 \: - \: A) \: = \: \sin A}} \\ \\ \implies {\sf{\dfrac{\sin A}{\cos A} \: + \: \dfrac{\cos A}{\sin A}}} \\ \\ \implies {\sf{\dfrac{\sin A \: \times \: \sin A \: + \: \cos A \: \times \: \cos A}{\sin A \cos A}}} \\ \\ \implies {\sf{\dfrac{\sin^2A \: + \: \cos ^2 A}{\sin A \cos A}}} \\ \\ \footnotesize {\tt{\dag \: \: \: \: \: \: \: \sin ^2A \: + \: \cos ^2 A \: = \: 1}} \\ \\ \implies {\sf{\dfrac{1}{\sin A \cos A}}} \\ \\ \implies {\sf{\dfrac{1}{\sin A} . \dfrac{1}{\cos A}}} \\ \\ \footnotesize\tt{\dag \: \: \: \: \: \dfrac{1}{\cos A} \: = \: \sec A \: \: \: \: and \: \: \: \: \dfrac{1}{\sin A} \: = \: \csc A}$

$\implies {\sf{\csc A \sec A}}$

$\mathbb{L.H.S \: = \: R.H.S}$

Here Proved