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UPSC CDS 1 2022 Math Paper

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Question : 87 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: 👈: Video Solution

Given:

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

Formula Used:

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

Calculation:

We have

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

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

\Rightarrow \frac{\cos \theta \csc \theta-\cos \theta+\cos \theta \csc \theta+\cos \theta}{\cot^{2}\theta}=2

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

\Rightarrow \cos \theta \times \left(\frac{1}{\sin \theta}\right)=\cot^{2}\theta

\Rightarrow \cot \theta = \cot^{2}\theta

\Rightarrow \cot \theta = 1

\Rightarrow \theta = 45^{\circ}

Now, We have to find the value of

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

Putting

\theta = 45^{\circ}

, we get

\Rightarrow \sin^{4}45^{\circ}+\cos^{4}45^{\circ}

\Rightarrow \left(\frac{1}{4}\right)+\left(\frac{1}{4}\right)=\frac{1}{2}

\therefore

The value of

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

is

\frac{1}{2}

.

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