AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

It is assumed x=tan‌θ, then we have
a‌tan‌θ+b‌sec‌θ=c
a‌sin‌θ+b=c‌cos‌θ
cos‌θ−a‌sin‌θ=b
Let, a2+c2=r2
c=r‌cos‌α,a=r‌sin‌α
tan‌α=

a

c

cos‌α×cos‌θ−sin‌α×sin‌θ=

b

r

Thus, cos(α+θ)=

b

r

α+θ=±cos−1

b

r

Let, a+θ be the positive solution and a+ϕ the negative solution, where
y=tan‌ϕ
α+ϕ=−(α+θ)
−2α=θ+ϕ
tan(−2α)=tan(θ+ϕ)
On expanding the above expression we get.

−2‌tan‌α

1−tan2α

=tan(θ+ϕ)
Substitute the values in above expression.

−2a c

1−( a c )2

=

x+y

1−xy

2ac

a2−c2

=

x+y

1−xy