Question

Prove that:
(1 + cot A + tan A)(sin A -cos A)/sec^3A-cosec^3A = sin^2A.cos^2A.

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Answer 1

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Answer 2

$\sf \mathfrak{\huge \underline{To~Prove:}}$

$\sf \frac {(1 + cot A + tan A)( Sin A - Cos A)} {sec^3 A -cosec^3 A} =sin^2A cos^2A$

$\mathfrak{\huge \underline{Solution:}}$

$\sf \frac {(1 + cot A + tan A)( Sin A - Cos A)} {sec^3 A -cosec^3 A} =sin^2A cos^2A \\\\\sf \longrightarrow\frac {(1+\frac{cosA} {sinA}+\frac{sinA}{cosA}) (SinA-CosA)}{\frac{1}{cos^3A}-\frac{1}{sin^3A}} \\\\\sf\longrightarrow \frac{(\frac{sinAcosA+cos^2A+sin^2A} {sinAcosA})(sinA-cosA)}{\frac{sin^3A-cos^3A}{cos^3Asin^3A}} \\\\\sf\longrightarrow \frac{(sinAcosA+1) (sinA-cosA)}{sinAcosA} \times \frac{cos^3Asin^3A}{sin^3A-cos^3A} \\\\\sf\longrightarrow \frac{(sinAcosA+1) (sinA-cosA)}{sinAcosA} \times \frac{cosAsinA(cos^2Asin^2A)}{sin^3A-cos^3A} \\\\\sf \longrightarrow\frac{(sinAcosA+1)(sinA-cosA)}{\cancel{sinAcosA}} \times \frac{\cancel{cosAsinA}(cos^2Asin^2A)}{sin^3A-cos^3A} \\\\\sf \longrightarrow \frac{(sinAcosA+1)(sinA-cosA)(cos^2Asin^2A)}{sin^3A-cos^3A}\\\\\sf By~identity \\\\\sf \boxed{a^3-b^3=(a-b)(a^2+b^2+ab)} \\\\\sf sin^3A-cos^3A= (sinA-cosA) (sin^2A+cos^2A+sin^3Acos^3A) \\\\\sf so, further~ simplification ~is~as~follows: \\\\\sf \longrightarrow \frac{(sinAcosA+1)(sinA-cosA)(cos^2Asin^2A)}{(sinA-cosA) (sin^2A+cos^2A+sin^3Acos^3A)}\\\\\sf \longrightarrow \frac{(sinAcosA+1)\cancel {(sinA-cosA)} (cos^2Asin^2A)}{\cancel {(sinA-cosA)} (sin^2A+cos^2A+sinAcosA)}\\\\\sf \longrightarrow \frac{(sinAcosA+1)(cos^2Asin^2A)}{1+sinAcosA} \\\\\sf \longrightarrow \frac{(\cancel{sinAcosA+1})(cos^2Asin^2A)}{\cancel{1+sinAcosA}} \\\\\sf \longrightarrow \huge\boxed{cos^2Asin^2A}$