What is cosθ /1-tanθ +sinθ /1-cotθ equal to?

Correct Answer is : sinθ+cosθsin θ + cos θsinθ+cosθ

Correct Answer is : sinθ+cosθsin θ + cos θsinθ+cosθ

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UPSC NDA I 2015 Mathematics Paper

Question : 39 of 120

Marks: +1, -0

What is

\frac{cos\theta }{1-tan\theta }+\frac{sin\theta }{1-cot\theta }

equal to?

Solution:

Concept:

Simplify the expression by converting

\tan\theta

and

\cot\theta

into

\frac{\sin\theta}{\cos\theta}

and

\frac{\cos\theta}{\sin\theta}

respectively, then combine terms.

Explanation:

First, rewrite

\tan\theta

and

\cot\theta

\frac{\cos\theta}{1 - \tan\theta} + \frac{\sin\theta}{1 - \cot\theta} < br>= \frac{\cos\theta}{1 - \frac{\sin\theta}{\cos\theta}} + \frac{\sin\theta}{1 - \frac{\cos\theta}{\sin\theta}}

Simplify the denominators:

= \frac{\cos\theta}{\frac{\cos\theta - \sin\theta}{\cos\theta}} + \frac{\sin\theta}{\frac{\sin\theta - \cos\theta}{\sin\theta}}

= \frac{\cos^2\theta}{\cos\theta - \sin\theta} + \frac{\sin^2\theta}{\sin\theta - \cos\theta}

Notice that

\sin\theta - \cos\theta = -(\cos\theta - \sin\theta)

. So the second term becomes:

= \frac{\cos^2\theta}{\cos\theta - \sin\theta} - \frac{\sin^2\theta}{\cos\theta - \sin\theta}

Combine the fractions:

= \frac{\cos^2\theta - \sin^2\theta}{\cos\theta - \sin\theta}

Factor the numerator using

a^2 - b^2 = (a-b)(a+b)

= \frac{(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)}{\cos\theta - \sin\theta}

Cancel the common factor

\cos\theta - \sin\theta

(provided

\cos\theta \neq \sin\theta

= \cos\theta + \sin\theta

Thus, the expression simplifies to

\sin\theta + \cos\theta

Answer:

\sin\theta + \cos\theta

(Option B)

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