Test Index

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 : 42 of 120

Marks: +1, -0

What is (p+q) equal to?

Solution:

We are given:

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

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

We need to find

p+q

p+q=\left(\frac{\sin^2(2\alpha)-\cos^2(2\alpha)}{\sin(2\alpha)\cdot\cos(2\alpha)}\right)+\left(\frac{\cos^2(\alpha)-\sin^2(\alpha)}{\sin(\alpha)\cdot\cos(\alpha)}\right)

Simplifying both terms:

=\frac{-2\cos(4\alpha)}{\sin(4\alpha)}+\frac{2\cos(2\alpha)}{\sin(2\alpha)}

Now, factorizing and simplifying further:

=\frac{2\left(\sin(4\alpha)\cdot\cos(2\alpha)-\cos(4\alpha)\cdot\sin(2\alpha)\right)}{\sin(4\alpha)\cdot\sin(2\alpha)}

Recognizing the sine identity, we get:

=\frac{2\sin(4\alpha-2\alpha)}{\sin(4\alpha)\cdot\sin(2\alpha)}

Finally, simplifying this:

=\frac{2\sin(2\alpha)}{\sin(4\alpha)\cdot\sin(2\alpha)}

The final result is:

p+q=2\csc(4\alpha)

Go to Question:

Prev Question

Next Question