Question

prove that
cos(90-theta)*sin(90-theta)/tan(90-theta)=sin^theta

Answer 1

Solution :

Here I am using A instead of theta.

LHS = [cos(90-A)sin(90-A)]/tan(90-A)

= ( sinAcosA)/cotA

= ( sinAcosA )/( cosA/sinA )

= ( sinAcosA × sinA )/cosA

After cancellation, we get

= sin²A

= RHS

••••

Answer 2

...............cos(90+x)=-sinx, where 90 and x are both measured in degrees.

L.H.S. = cos(90+x)= cos(180–90+x)=cos[180–(90-x)]=-cos(90-x)=-sin x

This transformation process makes use of two trigonometric identities,

cos(180-x)=-cosxcos(90-x)=sinx

There are also similar identities, and they can be conveniently remembered by using the concept of quadrants (“CAST”):