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

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!

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) = $Prove that (cosecA – sinA) (secA-cosA)$

Ans: Let’s start with LHS:

(cosec A – sin A) (sec A- cos A)

= ( $Prove that (cosecA – sinA) (secA-cosA)$ – $Prove that (cosecA – sinA) (secA-cosA)$ A) ( $Prove that (cosecA – sinA) (secA-cosA)$ – $Prove that (cosecA – sinA) (secA-cosA)$ A)