Question

Prove that

1+cotA+tanA(sinA-cosA) / sec3 A  - cosec3  A   =   sin2Acos2A       ​

Answer 1

ANSWER:

$\sf \displaystyle \frac{(1+cotA+tanA)(sinA-cosA)}{sec^3A-cosec^3A} =sin^2Acos^2A$

▣ Sec³A-cosec³A is in the form a³- b³=(a - b)(a² + b² + ab)

$\to \sf \displaystyle \frac{(1+cotA+tanA)(sinA-cosecA)}{(secA-cosecA)(sec^2A+cosec^2A+secAcosecA)}$

▣ Write all in terms of sin and cos.

$\to \sf \displaystyle \frac{(1+\dfrac{cosA}{sinA} +\dfrac{sinA}{cosA} )(sinA-cosA)}{(\dfrac{1}{cosA} -\dfrac{1}{sinA} )(sec^2A+cosec^2A+secAcosecA)}$

Further simplifying, we get:

$\to \sf \displaystyle \frac{(\dfrac{sinAcosA+cos^2A+sin^2A}{sinAcosA} )(sinA-cosA)}{(\dfrac{sinA-cosA}{cosAsinA} )(sec^2A+cosec^2A+secAcosecA)}$

▣ Divide the whole equation by sinAcosA on the num and denominator.

$\to \sf \displaystyle \frac{\dfrac{(\dfrac{sinAcosA+cos^2A+sin^2A}{sinAcosA} )(sinA-cosA)}{sinAcosA}}{\dfrac{(\dfrac{sinA-cosA}{cosAsinA} )(sec^2A+cosec^2A+secAcosecA)}{sinAcosA}}$

▣ Further simplifying, we get:

$\sf \displaystyle \frac{(sin^2A+cos^2A+sinAcosA)}{(sec^2A+cosec^2A+secAcosecA) }$

$\sf \displaystyle \frac{(1+sinAcosA)}{(\dfrac{1}{sin^2A} +\dfrac{1}{cos^2A} +secAcosecA) }$

$\sf \displaystyle \frac{(1+sinAcosA)}{\dfrac{cos^2A+sin^2A}{cos^2Asin^2A} +secAcosecA) }$

$\sf \displaystyle \frac{(1+sinAcosA)}{\dfrac{1}{cos^2Asin^2A} +secAcosecA }$

$\sf \displaystyle \frac{(1+sinAcosA)}{\dfrac{1}{cos^2Asin^2A} +\dfrac{1}{sinAcosA} }$

$\sf \displaystyle \frac{(1+sinAcosA)}{\dfrac{sinAcosA+cos^2Asin^2A}{cos^3Asin^3A} }$

$\sf \displaystyle \frac{(1+sinAcosA)}{\dfrac{sinAcosA(1+cosAsinA)}{cos^3Asin^3A} }$

$\sf \displaystyle \frac{1}{\dfrac{sinAcosA}{cos^3Asin^3A} }$

$\sf \dfrac{sin^3Acos^3A}{sinAcosA}$

$\to \boxed{\sf{\green{sin^2Acos^2A}}}$

∴LHS = RHS.

Hence proved.