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UPSC NDA I 2021 Mathematics Solved Paper

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Question : 40 of 120

Marks: +1, -0

What is

\frac{1+\tan^2\theta}{1+\cot^2\theta} - \left( \frac{1-\tan\theta}{1-\cot\theta} \right)^2

equal to?

0
1
$2 \tan \theta$
$2 \cot \theta$

Solution:

Concept:

Use trigonometric identities

\sec^{2}\theta = 1+\tan^{2}\theta

and

\csc^{2}\theta = 1+\cot^{2}\theta

to simplify the expression.

Explanation:

First, simplify the first term:

\frac{1+\tan^{2}\theta}{1+\cot^{2}\theta} = \frac{\sec^{2}\theta}{\csc^{2}\theta} = \frac{1/\cos^{2}\theta}{1/\sin^{2}\theta} = \frac{\sin^{2}\theta}{\cos^{2}\theta} = \tan^{2}\theta

.

For the second term, express

\cot\theta

as

\frac{1}{\tan\theta}

:

\frac{1-\tan\theta}{1-\frac{1}{\tan\theta}} = \frac{1-\tan\theta}{\frac{\tan\theta-1}{\tan\theta}} = (1-\tan\theta) \cdot \frac{\tan\theta}{\tan\theta-1}

.

Since

\tan\theta-1 = -(1-\tan\theta)

, this becomes

(1-\tan\theta) \cdot \frac{\tan\theta}{-(1-\tan\theta)} = -\tan\theta

.

Thus

\left(\frac{1-\tan\theta}{1-\cot\theta}\right)^2 = (-\tan\theta)^2 = \tan^{2}\theta

.

Now subtract:

\tan^{2}\theta - \tan^{2}\theta = 0

.

Answer:

0

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