Prove that (cosec A – sin A) (sec A-cos A) = 1/(cot A+tan A)

Q) Prove that (cosec A – sin A) (sec A- cos A) = $\frac{1}{(\cot A + \tan A)}$

Ans: Let’s start with LHS:

(cosec A – sin A) (sec A- cos A)

= ( $\frac{1}{\sin A}$ – $\sin$ A) ( $\frac{1}{\cos A}$ – $\cos$ A)

= ( $\frac{1 - \sin^2 A}{\sin A}$ ) ( $\frac{1 - \cos^2 A}{\cos A}$ )

= ( $\frac{\cos^2 A}{\sin A}$ ) ( $\frac{\sin^2 A}{\cos A}$ )

= cos A . sin A

Now, Let’s solve RHS:

$\frac{1}{(\cot A + \tan A)}$

= $\frac{1}{\frac{\cos A}{\sin A} + \frac{\sin A}{\cos A}}$

= $\frac{\sin A \cos A}{\cos^2 A + \sin^2 A}$

= sin A . cos A ………… same as LHS…. Hence Proved!

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