# Concept of trigonometric function

**Question:**

If !!tan(alpha/2)=m!! , then

!!(1-2sin^2(alpha/2))/(1-sinalpha)=?!!

(a) !!(1-m)/(2m)!!

(b) !! (2m)/(1+m)!!

(c) !!(1-m)/(1+m)!!

(d) !!(1+m)/(1-m)!!

**Solution**

Given that, !!tan(alpha/2)=m!!

!!f(alpha)= (1-2sin^2(alpha/2))/(1-sinalpha)!!

We know that, !!cos2A = 1-2sin^2A!!,

!!cos2A = (1-tan^2A)/(1+tan^2A)!! and

!!sin2A = (2tanA)/(1+tan^2A)!!

!!f(alpha)= cosalpha/(1+sinalpha)!!

Put the value of !! sin and cos!! in terms of !!tan!!

!! f(alpha)= ((1-tan^2(alpha/2))/(1+tan^2(alpha/2)))/(1+(2tan(alpha/2))/(1+tan^2(alpha/2)))!!

!!=> f(alpha)=(1-tan^2(alpha/2))/(1+tan^2(alpha/2)+2tan(alpha/2))!!

Put the value of !!tan(alpha/2)!!

!!=>f(alpha)= (1-m^2)/(1+m^2+2m) = ((1-m)(1+m))/(1+m)^2!!

!!=> (1-m)/(1+m)!!

Therefore, option (c) is correct.