Question

Question #12d43

Answer 1

See proof below.

Explanation:

Let us assume that `theta/2=A`
Our question becomes
Given `t=tanA`, to prove `(1−sin2A)/(cos2A)=(1-t)/(1+t)`
(this step is not essential as we could have written `theta=2 theta/2` and continued with the solution).

RHS is
`(1−sin2A)/(cos2A)`
Using expansion of double angle formula we get
`(1−2sinAcosA)/(cos^2A−sin^2A)`
Multiply numerator and denominator by `sec^2A`, we get

`(sec^2A−2tanA)/(1−tan^2A)`
Substituting `sec^2A=1+tan^2A`, we get
`(1+tan^2A−2tanA)/(1−tan^2A)`
`=>(1−tanA)^2/((1−tanA)(1+tanA))`
`=>(1−tanA)/(1+tanA)`
Substituting the given value of `tan A=t` we get
`(1−t)/(1+t)=RHS`
Proved.