Question

HØLA Guys❗

Solution please

Answer 1

Step-by-step explanation:

$Given:\frac{cot^3Asin^3A}{(cosA+sinA)^2}+\frac{tan^3Acos^3A}{(cosA+sinA)^2}$

$=\frac{\frac{cos^3A}{sin^3A}*sin^3A}{(sinA+cosA)^2}+\frac{\frac{sin^3A}{cos^3A}*cos^3A}{(cosA+sinA)^2}$

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

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

$=\frac{(cosA+sinA)(cos^2A+sin^2A-cosAsinA)}{(cosA+sinA)^2}$

$=\frac{(cos^2A+sin^2A-cosAsinA)}{cosA+sinA}$

$=\frac{1-cosAsinA}{cosA+sinA}$

$=\frac{1-\frac{1}{secAcosecA}}{\frac{1}{secA}+\frac{1}{cosecA}}$

$=\frac{\frac{secAcosecA-1}{secAcosecA}}{\frac{secA+cosecA}{secAcosecA}}$

$=\frac{secAcosecA-1}{secA+cosecA}$

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