Question

Evaluate the following :

`((sin 27^@)/(cos 63^@))^2 - (cos 63^@/sin 27^@)^2`

Answer 1

 We have to find: `((sin 27^@)/(cos 63^@))^2 - (cos 63^@/sin 27^@)^2`

Since `sin(90^@ - theta) = cos theta` and `cos(90^@ -  theta) = sintheta`

`(sin 27^@/cos 63^@)^2 - (cos 63^@/sin 27^@)^2 = (sin (90^@ - 63^@)/cos 63^@)^2 - ((cos (90^@ - 27^2))/sin 27^@)^2`

`= ((cos 63^@)/(cos 63^@))^2 - ((sin 27)/sin 27^@)^2`

 = 1 -1

= 0

So value of `((sin 27^@)/cos 63^@)^2 - ((cos 63^@)/(sin 27^@))^2` is 0