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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

If ( (x+y))/( (x-y)) = (a+b)/(a-b) , where a, b ≠ 0 and x, y are...

If sin⁡(x+y)sin⁡(x−y)=a+ba−b\dfrac{\sin(x+y)}{\sin(x-y)} = \dfrac{a+b}{a-b}, where a,b≠0a, b \neq 0 and x,yx, y are such that tan⁡x\tan x and tan⁡y\tan y are well-defined and non-zero, then which of the following relations is true?

Options

  1. Aatan⁡x=btan⁡ya \tan x = b \tan y
  2. Batan⁡x+btan⁡y=0a \tan x + b \tan y = 0
  3. Cbtan⁡x+atan⁡y=0b \tan x + a \tan y = 0
  4. Dbtan⁡x=atan⁡yb \tan x = a \tan y

Correct answer

D. btan⁡x=atan⁡yb \tan x = a \tan y

Step-by-step solution

Given sin⁡(x+y)sin⁡(x−y)=a+ba−b\dfrac{\sin(x+y)}{\sin(x-y)} = \dfrac{a+b}{a-b}

Applying componendo and dividendo, we get:

sin⁡(x+y)+sin⁡(x−y)sin⁡(x+y)−sin⁡(x−y)=(a+b)+(a−b)(a+b)−(a−b)\dfrac{\sin(x+y)+\sin(x-y)}{\sin(x+y)-\sin(x-y)} = \dfrac{(a+b)+(a-b)}{(a+b)-(a-b)}

Using the sum and difference formulas for sine:

2sin⁡xcos⁡y2cos⁡xsin⁡y=2a2b\dfrac{2\sin x \cos y}{2\cos x \sin y} = \dfrac{2a}{2b}

⇒tan⁡xtan⁡y=ab\Rightarrow \dfrac{\tan x}{\tan y} = \dfrac{a}{b}

⇒btan⁡x=atan⁡y\Rightarrow b \tan x = a \tan y

Answer: btan⁡x=atan⁡yb \tan x = a \tan y
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