Question

prove (cosecA-sinA)secA-cosA=1/tanA+cotA​

Answer 1

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Answer 2

Answer:

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

Consider,

LHS = $(cosec\,A-sin\,A)(sec\,A-cos\,A)$

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

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

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

       $=cos\,A\:\:sin\,A$

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.