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

JEE Main · Mathematics · Trigonometric Ratios & Identities

If (y+z-x), (z+x-y) and (x+y-z) are in A.P., then

If sin⁡(y+z−x),sin⁡(z+x−y)\sin (y+z-x), \sin (z+x-y) and sin⁡(x+y−z)\sin (x+y-z) are in A.P., then

Options

  1. A2tan⁡y=tan⁡x−tan⁡z2 \tan y=\tan x-\tan z
  2. Btan⁡y=tan⁡x+tan⁡z\tan y=\tan x+\tan z
  3. C2tan⁡y=tan⁡x+tan⁡z2 \tan y=\tan x+\tan z
  4. Dtan⁡y=tan⁡x−tan⁡z\tan y=\tan x-\tan z

Correct answer

C. 2tan⁡y=tan⁡x+tan⁡z2 \tan y=\tan x+\tan z

Step-by-step solution

As sin⁡(y+z−x),sin⁡(z+x−y)\sin (y+z-x), \sin (z+x-y) and sin⁡(x+y−z)\sin (x+y-z) are in A.P. ∴sin⁡(z+x−y)−sin⁡(y+z−x)=sin⁡(x+y−z)−sin⁡(z+x−y)⇒2cos⁡zsin⁡(x−y)=2cos⁡xsin⁡(y−z)⇒cos⁡zsin⁡(x−y)=cos⁡xsin⁡(y−z)⇒cos⁡zsin⁡(x−y)cos⁡xcos⁡ycos⁡z=cos⁡xsin⁡(y−z)cos⁡xcos⁡ycos⁡z⇒sin⁡(x−y)cos⁡xcos⁡y=sin⁡(y−z)cos⁡ycos⁡z\therefore \sin (z+x-y)-\sin (y+z-x)=\sin (x+y-z)-\sin (z+x-y)\Rightarrow 2 \cos z \sin (x-y)=2 \cos x \sin (y-z)\Rightarrow \cos z \sin (x-y)=\cos x \sin (y-z)\Rightarrow \frac{\cos z \sin (x-y)}{\cos x \cos y \cos z}=\frac{\cos x \sin (y-z)}{\cos x \cos y \cos z}\Rightarrow \frac{\sin (x-y)}{\cos x \cos y}=\frac{\sin (y-z)}{\cos y \cos z} Using sin⁡(A−B)cos⁡Acos⁡B=tan⁡A−tan⁡B⇒tan⁡x−tan⁡y=tan⁡y−taz⁡⇒tan⁡y−tan⁡x=tan⁡z−tan⁡y∴tan⁡x,tan⁡yand⁡tan⁡zare⁡\frac{\sin (A-B)}{\cos A \cos B}=\tan A-\tan B\Rightarrow \tan x-\tan y=\tan y-\operatorname{taz}\Rightarrow \tan y-\tan x=\tan z-\tan y\therefore \tan x, \tan y \operatorname{and} \tan z \operatorname{are} in A.P.
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