Question

Cot3A*sin3A/(cosA+sinA)2 +tan3A*cos3A/(cosA+sinA)2=secA*cosecA-1/cosrcA+secA

Answer 1

$\text{consider,}$

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

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

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

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

$\text{Using,}$

$\boxed{\bf\;a^3+b^3=(a+b)(a^2ab+b^2)}$

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

$=\displaystyle\frac{cos^2A-cosA\;sinA+sin^2A}{cosA+sinA}$

$=\displaystyle\frac{1-cosA\;sinA}{cosA+sinA}$

$=\displaystyle\frac{1-(\frac{1}{secA})(\frac{1}{cosecA})}{\frac{1}{secA}+\frac{1}{cosecA}}$

$=\displaystyle\frac{1-\frac{1}{secA\;cosecA}}{\frac{cosecA+secA}{secA\;cosecA}}$

$=\displaystyle\frac{\frac{secA\;cosecA-1}{secA\;cosecA}}{\frac{cosecA+secA}{secA\;cosecA}}$

$=\displaystyle\frac{secA\;cosecA-1}{cosecA+secA}$

$\implies\boxed{\bf\frac{cot^3A*sin^3A}{(cosA+sinA)^2}+\frac{tan^3A*cos^3A}{(cosA+sinA)^2}=\frac{secA\;cosecA-1}{cosecA+secA}}$

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7 cosecA -3 cotA =7 prove that 7 cotA - 3 cosecA =3

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Answer 2

Answer:

\text{consider,}consider,

\displaystyle\frac{cot^3A*sin^3A}{(cosA+sinA)^2}+\frac{tan^3A*cos^3A}{(cosA+sinA)^2}(cosA+sinA)2cot3A∗sin3A+(cosA+sinA)2tan3A∗cos3A

=\displaystyle\frac{\frac{cos^3A}{sin^3A}*sin^3A}{(cosA+sinA)^2}+\frac{\frac{sin^3A}{cos^3A}*cos^3A}{(cosA+sinA)^2}=(cosA+sinA)2sin3Acos3A∗sin3A+(cosA+sinA)2cos3Asin3A∗cos3A

=\displaystyle\frac{cos^3A}{(cosA+sinA)^2}+\frac{sin^3A}{(cosA+sinA)^2}=(cosA+sinA)2cos3A+(cosA+sinA)2sin3A

=\displaystyle\frac{cos^3A+sin^3A}{(cosA+sinA)^2}=(cosA+sinA)2cos3A+sin3A

\text{Using,}Using,

\boxed{\bf\;a^3+b^3=(a+b)(a^2ab+b^2)}a3+b3=(a+b)(a2ab+b2)

=\displaystyle\frac{(cosA+sinA)(cos^2A-cosA\;sinA+sin^2A)}{(cosA+sinA)^2}=(cosA+sinA)2(cosA+sinA)(cos2A−cosAsinA+sin2A)

=\displaystyle\frac{cos^2A-cosA\;sinA+sin^2A}{cosA+sinA}=cosA+sinAcos2A−cosAsinA+sin2A

=\displaystyle\frac{1-cosA\;sinA}{cosA+sinA}=cosA+sinA1−cosAsinA

=\displaystyle\frac{1-(\frac{1}{secA})(\frac{1}{cosecA})}{\frac{1}{secA}+\frac{1}{cosecA}}=secA1+cosecA11−(secA1)(cosecA1)

=\displaystyle\frac{1-\frac{1}{secA\;cosecA}}{\frac{cosecA+secA}{secA\;cosecA}}=secAcosecAcosecA+secA1−secAcosecA1

=\displaystyle\frac{\frac{secA\;cosecA-1}{secA\;cosecA}}{\frac{cosecA+secA}{secA\;cosecA}}=secAcosecAcosecA+secAsecAcosecAsecAcosecA−1

=\displaystyle\frac{secA\;cosecA-1}{cosecA+secA}=