Question

(Cosec a- sin a)(sec a- cos a)=1/tan a cot a

Answer 1

Hi ,

LHS = (cosecA - sinA)(SecA- cosA)

= (CosecA- 1/cosecA)(secA-1/secA)

= (Cosec²A-1)(sec²A-1)/(cosecAsecA)

= Cot² Atan²A / cosecAsecA

= 1/cosecAsecA

= SinAcosA

= RHS

I hope this helps you.

:)

Answer 2

Answer and explanation:

To prove : $(\csc a-\sin a)(\sec a-\cos a)=\frac{1}{\tan a+\cot a}$

Proof :

Taking LHS,

$LHS=(\csc a-\sin a)(\sec a-\cos a)$

$LHS=(\frac{1}{\sin a}-\sin a)(\frac{1}{\cos a}-\cos a)$

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

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

$LHS=\sin a\cos a$

Taking RHS,

$RHS=\frac{1}{\tan a+\cot a}$

$RHS=\frac{1}{\frac{\sin a}{\cos a}+\frac{\cos a}{\sin a}}$

$RHS=\frac{1}{\frac{\sin^2 a+\cos^2 a}{\sin a\cos a}}$

$RHS=\frac{1}{\frac{1}{\sin a\cos a}}$

$RHS=\sin a\cos a$

LHS = RHS

Hence proved.