Question

If tan y=3tan x ,prove that tan(x+y)=2sin 2y/1+2 cos y

Answer 1

Answer:

Given,

tan y = 3 tan x

$\implies \tan x = \frac{\tan y}{3}-----(1)$

We have to prove that,

$\tan (x+y) = \frac{2\sin 2y}{1+2\cos 2y}$

L.H.S.

$=\tan (x+y)$

$=\frac{\tan x+\tan y}{1-\tan x\tan y}$

$=\frac{\frac{\tan y}{3}+\tan y}{1-\frac{\tan y}{3}\tan y}$

( From equation (1) )

$=\frac{\frac{4\tan y}{3}}{\frac{3-\tan^2 y}{3}}$

$=\frac{4\tan y}{3-\tan^2 y}$

$=\frac{4\frac{\sin y}{\cos y}}{3-\frac{\sin^2 y}{\cos^2y}}$

$=\frac{4\sin y\cos y}{3\cos^2 y-\sin^2 y}$

$=\frac{2\sin 2y}{2\cos^2 y + \cos^2 y -\sin^2 y}$

( ∵ sin 2x = 2 sinx cosx )

$=\frac{2\sin 2y}{1+\cos 2y+\cos 2y}$

( cos 2x = cos² x - sin² x = 2cos² x - 1 )

$=\frac{2\sin 2y}{1+2\cos 2y}$

= R.H.S.

Hence, proved....

Answer 2

Step-by-step explanation:

Given,

tan y = 3 tan x

\implies \tan x = \frac{\tan y}{3}-----(1)⟹tanx=

3

tany

−−−−−(1)

We have to prove that,

\tan (x+y) = \frac{2\sin 2y}{1+2\cos 2y}tan(x+y)=

1+2cos2y

2sin2y

L.H.S.

=\tan (x+y)=tan(x+y)

=\frac{\tan x+\tan y}{1-\tan x\tan y}=

1−tanxtany

tanx+tany

=\frac{\frac{\tan y}{3}+\tan y}{1-\frac{\tan y}{3}\tan y}=

1−

3

tany

tany

3

tany

+tany

( From equation (1) )

=\frac{\frac{4\tan y}{3}}{\frac{3-\tan^2 y}{3}}=

3

3−tan

2

y

3

4tany

=\frac{4\tan y}{3-\tan^2 y}=

3−tan

2

y

4tany

=\frac{4\frac{\sin y}{\cos y}}{3-\frac{\sin^2 y}{\cos^2y}}=

3−

cos

2

y

sin

2

y

4

cosy

siny

=\frac{4\sin y\cos y}{3\cos^2 y-\sin^2 y}=

3cos

2

y−sin

2

y

4sinycosy

=\frac{2\sin 2y}{2\cos^2 y + \cos^2 y -\sin^2 y}=

2cos

2

y+cos

2

y−sin

2

y

2sin2y

( ∵ sin 2x = 2 sinx cosx )

=\frac{2\sin 2y}{1+\cos 2y+\cos 2y}=

1+cos2y+cos2y

2sin2y

( cos 2x = cos² x - sin² x = 2cos² x - 1 )

=\frac{2\sin 2y}{1+2\cos 2y}=

1+2cos2y

2sin2y

= R.H.S.

Hence, proved....