Question

How do you simplify `f(theta)=cot(theta/4)-tan(theta/2+pi/2)` to trigonometric functions of a unit `theta`?

Answer 1

`f(theta)=cot(theta/4)-tan(theta/2+pi/2)=sqrt(2/(1-costheta))`

Explanation:

As `tan(90^@+A)=-cotA`, `tan(theta/2+pi/2)=-cot(theta/2)`

Hence `f(theta)=cot(theta/4)-tan(theta/2+pi/2)`

= `cot(theta/4)-cot(theta/2)`

Further `cot(A/2)=cos(A/2)/sin(A/2)=(2cos^2(A/2))/(2sin(A/2)cos(A/2)`

= `(1+cosA)/sinA=cscA+cotA`

Also `sin(A/2)=sqrt((2sin^2(A/2))/2)=sqrt((1-(1-2sin^2(A/2)))/2)`

= `sqrt((1-cosA)/2)` .......................(A)

Hence, `cot(theta/4)-cot(theta/2)`

= `csc(theta/2)+cot(theta/2)-cot(theta/2)`

= `csc(theta/2)`

= `1/sin(theta/2)`

= `sqrt(2/(1-costheta))` (using (A))