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UPSC NDA I 2025 Mathematics Paper

Question Numbers: 41-43

Consider the following for the three (03) items that follow:

Let p = tan 2α - tanα and q = cotα - cot 2α

Question : 43 of 120

Marks: +1, -0

What is tan²α equal to?

Solution:

We are given:

\;p=\tan(2\alpha)-\tan(\alpha)

\;q=\cot(\alpha)-\cot(2\alpha)

We need to find

\tan^2(\alpha)

in terms of

p

and

q

We start by simplifying the expression

\;\frac{p}{p+2q}

\;\frac{p}{p+2q} = \;\frac{1}{1+\;\frac{2q}{p}}

This simplifies further as:

\;\frac{1}{1+\;\frac{2}{\tan(\alpha)\cdot\tan(2\alpha)}}

Now, simplify the fraction inside the denominator:

=\;\frac{1}{1+\;\frac{2}{\tan(\alpha)\cdot\tan(2\alpha)}} = \;\frac{1}{1+\;\frac{1}{2}\cdot\tan(\alpha)\cdot\tan(2\alpha)}

Now, expand both terms using trigonometric identities:

=\;\frac{\sin(\alpha)\cdot\sin(2\alpha)}{\sin(\alpha)\cdot\cos(\alpha)+\cos(\alpha)\cdot\cos(2\alpha)}

Simplifying this expression gives:

=\tan^2(\alpha)

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