Question

Question #57b11

Answer 1

Use these trig identities:
`tan t = (2tan (t/2))/(1 - tan^2 (t/2))`
`sec t = 1/(cos t) = (1 + tan^2 (t/2))/(1 - tan^2 (t/2))`
Develop the Right Side of the equation:
`RS = (2tan (t/2) + 1 + tan^2 (t/2))/(1 - tan^2 (t/2)) =`
`RS = ((1 + tan(t/2))^2)/(1 - tan^2 (t/2))`

Divide both numerator and denominator by `(1 + tan (t/2))`, we get
`RS = (1 + tan(t/2))/(1 - tan (t/2)) = LS`

Answer 2

`LHS= (1 + tan(theta/2))/(1 - tan (theta/2))`

`= (cos(theta/2)(1 + tan(theta/2)))/(cos(theta/2)(1 - tan (theta/2)) )`

`= (cos(theta/2) + sin(theta/2))/(cos(theta/2) - sin (theta/2))`

Multiplying numerator and denominator by `(cos(theta/2) + sin(theta/2))` we get

`LHS= (cos(theta/2) + sin(theta/2))^2/(cos^2(theta/2) - sin ^2(theta/2))`

`= (cos^2(theta/2) + sin^2(theta/2)+2sin(theta/2)(costheta/2))/(cos^2(theta/2) - sin ^2(theta/2))`

`= (1+sin(theta))/cos(2xxtheta/2)`

`=1/costheta+sintheta/costheta`

`=tantheta+sectheta=RHS`

Proved