Question

Sina cosa/cosa(seca-coseca)*sin^2a-cos^2a/sin^3a+cos^3a=sina

Answer 1

Given: The correct expression is

1 - sinA cosA/cosA(secA - cosecA).sin^2A - cos^2A / sin^3A + cos^3A = sinA

To find: Prove the given expression.

Solution:

• Now we have given:

1 - sinA cosA/cosA(secA - cosecA).sin^2A - cos^2A / sin^3A + cos^3A = sinA

• Consider LHS, we have:

1 - sinA cosA/cosA(secA - cosecA).sin^2A - cos^2A / sin^3A + cos^3A

• Converting all terms to sin and cos, we get:

1 - sinA cosA/cosA(1/cosA - 1/sinA) x (sinA + cosA)(sinA - cosA)/(sinA + cosA)(sin^2A + cos^2A - sinAcosA)

• Now simplifying it, we get:

( 1 - sinA cosA) / cosA (sinA - cosA/cosAsinA) x (sinA - cosA)/(1 - sinAcosA)

• Simplifying it again we get:

               sin A

               RHS

• Hence proved.

Answer:

               So we have proved in solution part that

               1 - sinA cosA/cosA(secA - cosecA).sin^2A - cos^2A / sin^3A + cos^3A = sinA