Test Index

UPSC CDS 1 2022 Math Paper

© examsnet.com

Question : 60 of 100

Marks: +1, -0

If

\frac{\cos\theta}{\csc\theta+1}+\frac{\cos\theta}{\csc\theta-1}=2

0 < \theta<90^{\circ}

, then what is the value of

\sin^{4}\theta+\cos^{4}\theta

?

2
1
$\frac{1}{2}$
$\frac{1}{4}$

Solution:

Concept:

Simplify the given trigonometric equation to find the value of

\theta

, then substitute to evaluate

\sin^4\theta+\cos^4\theta

.

Explanation:

Start with

\displaystyle\frac{\cos\theta}{\csc\theta+1}+\frac{\cos\theta}{\csc\theta-1}=2

.

Take the least common denominator

(\csc\theta+1)(\csc\theta-1)=\csc^2\theta-1=\cot^2\theta

.

Combine the fractions:

\displaystyle\frac{\cos\theta(\csc\theta-1)+\cos\theta(\csc\theta+1)}{\cot^2\theta}=2

.

Simplify numerator:

\cos\theta\csc\theta-\cos\theta+\cos\theta\csc\theta+\cos\theta=2\cos\theta\csc\theta

.

Thus

\displaystyle\frac{2\cos\theta\csc\theta}{\cot^2\theta}=2

.

Cancel

2

on both sides:

\displaystyle\frac{\cos\theta\csc\theta}{\cot^2\theta}=1

.

Since

\cos\theta\csc\theta=\frac{\cos\theta}{\sin\theta}=\cot\theta

, we have

\displaystyle\frac{\cot\theta}{\cot^2\theta}=1

.

Simplify:

\displaystyle\frac{1}{\cot\theta}=1

, so

\cot\theta=1

.

For

0^{\circ} < \theta<90^{\circ}

,

\cot\theta=1

gives

\theta=45^{\circ}

.

Now compute

\sin^4\theta+\cos^4\theta

at

\theta=45^{\circ}

.

\sin45^{\circ}=\cos45^{\circ}=\frac{1}{\sqrt{2}}

.

Therefore

\sin^445^{\circ}=\left(\frac{1}{\sqrt{2}}\right)^4=\frac{1}{4}

, and

\cos^445^{\circ}=\frac{1}{4}

.

Sum:

\frac{1}{4}+\frac{1}{4}=\frac{1}{2}

.

Answer:

\displaystyle\frac{1}{2}

(option C).

© examsnet.com

Go to Question:

Prev Question

Next Question