Question

(cosecA -sinA) (secA-cosA)=1\tanA+cotA

Answer 1

Step-by-step explanation:

detailed explanation is shown above

Answer 2

Answer:

To  Prove: $(cosec\,A-sin\,A)(sec\,A-cos\,A)=\frac{1}{tan\,A+cot\,A}$

Consider,

LHS = ( cosec A - sin A )( sec A - cos A )

       = ( 1/sin A - sin A )( 1/cos A - cos A )

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

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

       $=cos\,A\:\:sin\,A$

RHS = $\frac{1}{tan\,A+cot\,A}$

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

       $=\frac{1}{\frac{sin^2\,A+cos^\,A}{cos\,A\:\:sin\,A}}$

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

       $=\frac{cos\,A\:\:sin\,A}{1}$

       $=cos\,A\:\:sin\,A$

LHS = RHS

Hence Proved.