UPSC NDA I 2024 Mathematics Paper

Firstly, we need to simplify the expression inside the square root: √15+cot2(‌

π

4

−2cot−13).
Let's start by simplifying the term inside the cotangent function: ‌

π

4

−2cot−13.
If we let x=cot−13, then by definition: cot‌x=3.
Thus, x=cot−13.
Now we need to find: cot(‌

π

4

−2x).
Using the cotangent subtraction formula:
cot(A−B)=‌

cot‌A‌cot‌B+1

cot‌B−cot‌A

Here, A=‌

π

4

and B=2x.
Since cot(‌

π

4

)=1, we have:
cot(‌

π

4

−2cot−13)=‌

1⋅cot(2x)+1

cot(2x)−1

Next, we need to find cot(2cot−13). Using the double-angle formula for cotangent: cot(2θ)=‌

cot2(θ)−1

2‌cot(θ)

.
In this case, θ=cot−13, so cot(θ)=3.
Then, we get:
cot(2θ)=‌

32−1

2⋅3

=‌

9−1

6

=‌

8

6

=‌

4

3

‌. ‌
Therefore, cot(2cot−13)=‌

4

3

.

cot(‌

π

4

−2cot−13)=‌

1⋅ 4 3 +1

4 3 −1

=‌

‌ 4 3 +1

4 3 −1

=‌

‌ 4 3 + 3 3

4 3 −‌ 3 3

=‌

‌ 7 3

1 3

=7‌. ‌
Finally, substituting this result into the original expression gives:
√15+cot2(‌

π

4

−2cot−13)=√15+72=√15+49=√64=8‌. ‌