Question

Prove that : sinA cosA -[cos(90-A)cos(90-A)cosA]/[sec(90-A)] - [sin(90-A)sin(90-A)sinA]/[cosec(90-A)] + sec(90-A)=1/sinA

Answer 1

step-by-step explanation:

Sina.cosa-sina cos(90-a).cosa/sec(90-a)-cosa.sin(90-a).sina/coseca(90-a)

=sina.cosa- sina sina .cosa/coseca-cosa.cosa.sina/seca

=sina.cosa-sina.sina.cosa.sina-cosa.cosa.sina.cosa

=sina.cosa-sin^3a.cosa-cos^3a.sina

=sina.cosa(1-sin^2a-cos^2a)

=sina.cosa(1-(sin^2a+cos^2a ))

=Sina.cosa(1-1)

=sina.cosa(0)

=0