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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2018 · Mathematics · Trigonometric Ratios & Identities

What is ( 5 x- 3 x)/( 5 x+ 3 x) equal to?

What is \frac{\sin 5 x-\sin 3 x}{\cos 5 x+\cos 3 x}$ equal to?

Options

  1. Asin x
  2. Bcos x
  3. Ctan x
  4. Dcot x

Correct answer

C. tan x

Step-by-step solution

sin⁡5x−sin⁡3xcos⁡5x+cos⁡3xsin⁡c−sin⁡d=2cos⁡(c+d2)sin⁡(c−d2)cos⁡c−cos⁡d=2cos⁡(c+d2)cos⁡(c−d2)∴sin⁡5x−sin⁡3xcos⁡5x+cos⁡3x=2cos⁡(5x+3x2)sin⁡(5x−3x2)2cos⁡(5x+3x2)cos⁡(5x−3x2)=sin⁡(2x2)cos⁡(2x2)=sin⁡xcos⁡x=tan⁡x\begin{aligned} & \frac{\sin 5 x-\sin 3 x}{\cos 5 x+\cos 3 x} \\ & \sin c-\sin d=2 \cos \left(\frac{c+d}{2}\right) \sin \left(\frac{c-d}{2}\right) \\ & \cos c-\cos d=2 \cos \left(\frac{c+d}{2}\right) \cos \left(\frac{c-d}{2}\right) \\ & \therefore \frac{\sin 5 x-\sin 3 x}{\cos 5 x+\cos 3 x}=\frac{2 \cos \left(\frac{5 x+3 x}{2}\right) \sin \left(\frac{5 x-3 x}{2}\right)}{2 \cos \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)} \\ & =\frac{\sin \left(\frac{2 x}{2}\right)}{\cos \left(\frac{2 x}{2}\right)}=\frac{\sin x}{\cos x}=\tan x\end{aligned}
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