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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2015 · Mathematics · Trigonometric Ratios & Identities

What is ( θ)/(1- θ)+( θ)/(1- θ) equal to?

What is \frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}$ equal to?

Options

  1. Asin⁡θ−cos⁡θ\sin \theta-\cos \theta
  2. Bsin⁡θ+cos⁡θ\sin \theta+\cos \theta
  3. C2sin⁡θ2 \sin \theta
  4. D2cos⁡θ2 \cos \theta

Correct answer

B. sin⁡θ+cos⁡θ\sin \theta+\cos \theta

Step-by-step solution

cos⁡θ1−tan⁡θ+sin⁡θ1−cot⁡θ=cos⁡θ1−sin⁡θcos⁡θ+sin⁡θ1−cos⁡θsin⁡θ=cos⁡2θcos⁡θ−sin⁡θ+sin⁡2θsin⁡θ−cos⁡θ\begin{aligned} & \frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta} \\ & =\frac{\cos \theta}{1-\frac{\sin \theta}{\cos \theta}}+\frac{\sin \theta}{1-\frac{\cos \theta}{\sin \theta}} \\ & =\frac{\cos ^2 \theta}{\cos \theta-\sin \theta}+\frac{\sin ^2 \theta}{\sin \theta-\cos \theta}\end{aligned}=cos⁡2θcos⁡θ−sin⁡θ−sin⁡2θcos⁡θ−sin⁡θ=cos⁡2θ−sin⁡2θcos⁡θ−sin⁡θ=(cos⁡θ−sin⁡θ)(cos⁡θ+sin⁡θ)cos⁡θ−sin⁡θ\begin{aligned} & =\frac{\cos ^2 \theta}{\cos \theta-\sin \theta}-\frac{\sin ^2 \theta}{\cos \theta-\sin \theta} \\ & =\frac{\cos ^2 \theta-\sin ^2 \theta}{\cos \theta-\sin \theta}=\frac{(\cos \theta-\sin \theta)(\cos \theta+\sin \theta)}{\cos \theta-\sin \theta}\end{aligned}\therefore \quad$ Option (B) is correct.
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