Question

How do you evaluate `cot 195`?

Answer 1

`cot195^o=2+sqrt3`

Explanation:

`cot195^o=cot(180^o +15^o)`

= `cot15^o`

Now `cot(x/2)=cos(x/2)/sin(x/2)`

= `(2cos(x/2)xxcos(x/2))/(2cos(x/2)sin(x/2))=(2cos^2(x/2))/sinx`

= `(1+cosx)/sinx=cscx+cotx`

Hence, `cot15^o=csc30^o +cot30^o=2+sqrt3`