AP EAPCET 25-Aug-2021 Shift 2 Paper

Given In a Δ‌ABC,tan(

A

2

,tan(

B

2

),tan(

C

2

) are in Arithmetic progression,
∴ tan(

B

2

)−tan(

A

2

)=tan(

C

2

)−tan(

B

2

)

sin( B 2 )

cos( B 2 )

−

sin( A 2 )

cos( A 2 )

=

sin( C 2 )

cos( C 2 )

−

sin( B 2 )

cos( B 2 )

sin( B 2 ).cos( A 2 )−sin( A 2 ).cos( B 2 )

cos( B 2 ).cos( A 2 )

=

sin( C 2 ).cos( B 2 )−sin( B 2 )‌cos( C 2 )

cos( C 2 ).cos( B 2 )

sin( B 2 − A 2 )

cos( A 2 )

=

sin( C 2 − B 2 )

cos( C 2 )

...(i)
∵ A+B+C=180° (angle sum property in Δ‌ABC)

A

2

=

1

2

{180−(B+C)}

A

2

=90−(

B

2

+

C

2

) and

C

2

=90−(

A

2

+

B

2

)
From Eq. (i), we get

sin( B 2 − A 2 )

cos{90−( B 2 + C 2 )}

=

sin( C 2 − B 2 )

cos{90−( A 2 + B 2 )}

⇒

sin( B 2 − A 2 )

sin( B 2 + C 2 )

=

sin( C 2 − B 2 )

sin( B 2 + A 2 )

⇒ sin(

B

2

+

A

2

).sin(

B

2

−

A

2

)
=sin(

C

2

+

B

2

).sin(

C

2

−

B

2

)
⇒ 2‌sin(

B

2

+

A

2

).sin(

B

2

−

A

2

)
=2‌sin(

C

2

+

B

2

).sin(

C

2

−

B

2

) {∵2‌sin‌x−sin‌y=cos(x−y)−cos(x+y)}
⇒ cos(

B

2

+

A

2

−

B

2

+

A

2

)−cos(

B

2

+

A

2

+

B

2

−

A

2

)
=cos(

C

2

+

B

2

−

C

2

+

B

2

)−cos(

C

2

+

B

2

+

C

2

−

B

2

)
cos‌A−cos‌B=cos‌B−cos‌C
⇒ 2‌cos‌B=cos‌A+cos‌C
⇒ cos‌A,cos‌B and cos‌C are in AP.