Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:$\text{(cosec A - sin A)(sec A - cos A)} =\frac{ 1}{\text{(tan A + cot A)}}$ [Hint : Simplify LHS and RHS separately]

Answer 1

$\text{(cosec A - sin A)(sec A - cos A)} = \frac{1}{\text{(tan A + cot A)}}$

L.H.S =$\text{ (cosec A - sin A)(sec A - cos A)}$

$⇒ (\frac{1}{\text{sin A }}-\text{ sin A)}(\frac{1}{\text{cos A }}- \text{cos A})$

$= \frac{\text{(1 - sin² A)}}{\text{sin A }}×\frac{\text{ (1 - cos² A)}}{\text{cos A}}$

$= \frac{\text{cos² A sin² A}}{\text{sin A cos A}}$

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

$=\frac{\text{ sin A cos A}}{\text{(sin² A + cos² A) }}$ [(sin² A + cos² A) = 1]

$= \frac{1}{\text{sin² A + cos² A}}$

$=\frac{ 1}{ [(\frac{\text{sin² A}}{\text{sin A cos A}}) + (\frac{\text{cos² A}}{\text{sin A cos A}})]}$ [ By dividing numerator and denominator by (sin A cos A)]

$= \frac{1}{[(\frac{\text{sin A}}{\text{cos A}}) + (\frac{\text{cos A}}{\text{sin A}})]}$

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

= RHS