Question

(1+cotA+tanA)(sinA-cosA)/sec^3A-cosec^3A=sin^2Acos^2A

Answer 1

I hope the answer is clear enough

Answer 2

Answer with Step-by-step explanation:

LHS

$\frac{(1+cotA+tanA)(sinA-cosA)}{sec^3A-cosec^3A}$

$\frac{(1+\frac{cosA}{sinA}+\frac{sinA}{cosA})(sinA-cosA)}{\frac{1}{cos^3A}-\frac{1}{sin^3A}}$

Using formula

$cotA=\frac{cosA}{sinA},tanA=\frac{sinA}{cosA},secA=\frac{1}{cosA},cosecA=\frac{1}{sinA}$

$\frac{(sinAcosA+cos^2A+sin^2)(sinA-cosA)}{sinAcosA(\frac{sin^3-cos^3}{sin^3Acos^3A})}$

$\frac{(sin^2A+cos^2A+sinAcosA)(sinA-cosA)}{sinAcosA(\frac{sin^3A-cos^3A}{sin^3Acos^3A})}$

$\frac{(sinAcosA+cos^2A+sin^2A)(sin^3Acos^3A)(sinA-cosA)}{sinAcosA(sinA-cosA)(sin^2A+cos^2A+sinAcosA)}$

By using identity:$a^3-b^3=(a-b)(a^2+b^2+ab)}$

$sin^2Acos^2A$

LHS=RHS

Hence, proved .

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