Question

Cot^3*sin^3/(cos+sin)^2+tan^3*cos^3/(cos+sin)^2

Answer 1

Answer:

prove that cot3 sin3/(cos+sin)2 +tan3 cos3/(cos-sin)2 =sec cosec-1/cosec+sec

Answer 2

$\sf \frac{cot^3 \theta sin^3 \theta} {(cos \theta+sin \theta)^2} +\frac{tan^3 \theta cos^3 \theta} {(cos \theta+sin \theta)^2} \\\\\sf </p><p>=\frac{cot^3 \theta sin^3 \theta+tan^3 \theta cos^3 \theta}{(cos \theta+sin \theta)^2} \\\\\sf </p><p>=\frac{\frac{cos^3 \theta}{sin^3 \theta} \: x \: sin^3 \theta+\frac{sin^3 \theta} {cos^3 \theta} \: x \: cos^3 \theta}{(cos \theta+sin \theta)^2} \\\\\sf </p><p>=\frac{\frac{cos^3 \theta}{\cancel{sin^3 \theta}} \: x \: \cancel{sin^3 \theta}+\frac{sin^3 \theta} {\cancel {cos^3 \theta}} \: x \: \cancel{cos^3 \theta}}{(cos \theta+sin \theta)^2} \\\\\sf </p><p>=\frac{cos^3 \theta + sin^3 \theta} {(cos \theta+sin \theta)^2} \\\\\sf </p><p>=\frac{(cos \theta+sin \theta)(cos^2 \theta+sin^2 \theta-cos \theta sin \theta)}{(cos \theta+sin \theta) ^2} \\\\\sf </p><p>=\frac{1-cos \theta sin \theta} {cos \theta+ sin \theta} </p><p>$