Question

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

Answer 1

correct question :

If tan y = 3 tan x, prove that tan (x + y) =

2 sin 2 y/1 + 2 cos 2 y

Given :

tan y = 3

To prove :

(x + y) = 2 sin 2 y/1 + 2 cos 2 y

Solution :

tan x = tan y/3

tan x + y = tan x + tan y/1 - tan x tan y

➸ (tan y/3) + tan y/1 - (tan² y/3)

➸ 4 tan y/3 - tan² y

➸ (4sin y/cos y)/(3 - sin² y/cos y)

➸ 4sin y cos y/3cos² y - sin² y

➸ 2 x 2sin y cos y/2cos² y + cos² y-sin²y

➸ 2sin 2y/1 + cos 2y + cos 2y

∴ 2sin 2y/1 + 2cos 2y ( hence proved )

Answer 2

Step-by-step explanation:

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

$\tan(x + y) = \frac{ \tan(x) + \tan(y) }{1 - \tan(x) \tan(y) } \\ = \frac{ \frac{1}{3} \tan(y) + \tan(y) }{1 - \frac{1}{3} \tan(y) \tan(y) } \\ \frac{ \frac{4}{3} \tan(y) }{ \frac{1}{ 3}(3 - { \tan }^{2} y) } \\ \frac{4 \frac{ \sin(y) }{ \cos(y) } }{3 - \frac{ { \sin }^{2}y }{\cos^{2}y } } \\ \frac{4 \frac{ \sin(y) }{ \cos(y) } }{ \frac{ 3 \cos^{2}y - sin^{2} y}{ \cos(^{2} ) } } \\ \frac{4 sin(y )cos(y )}{3 \cos^{2} y - (1 - \cos^{2} y) } \\ \frac{2 \sin(2y) }{4 \cos^{2}y - 1 } \\ \frac{2 \sin(2y) }{2(2 \cos^{2}y - 1) + 2 - 1 } \\ \ \frac{2 \sin(2y) }{2cos(2y )+ 1 }$