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UPSC CDS 1 2022 Math Paper

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Question : 14 of 100

Marks: +1, -0

If

\cos(x+y)=0

\sin(x-y)=\;\frac{1}{2}

x, y \in \left[0, \;\frac{\pi}{2}\right]

then what is the value of

\cot(2x-y)

?

0
$\;\frac{1}{2}$
1
2

Solution: 👈: Video Solution

Given:

\cos(x+y)=0

\sin(x-y)=\;\frac{1}{2}

Concept Used:

\sin A=\sin B

then

A=B

\cos A=\cos B

then

A=B

Where

0 \le A, B \le 90^{\circ}

Formula Used:

\cos 90^{\circ}=0, \sin 30^{\circ}=\frac{1}{2}, \cot 90^{\circ}=0

Calculation:

We have

\cos(x+y)=0

\Rightarrow \cos(x+y)=\cos 90^{\circ}

\Rightarrow \cos(x+y)=\cos 90^{\circ}

\Rightarrow x+y=90^{\circ}

And

\sin(x-y)=\;\frac{1}{2}

\Rightarrow \sin(x-y)=\sin 30^{\circ}

\Rightarrow x-y=30^{\circ}

On adding equation (i) and (ii), we get

2x=120^{\circ}

\Rightarrow x=60^{\circ}

Now, From (i)

60^{\circ}+y=90^{\circ}

\Rightarrow y=30^{\circ}

Now, We have to find

\cot(2x-y)

So,

\cot(2\times 60^{\circ}-30^{\circ})=\cot 90^{\circ}=0

\therefore

The value of

\cot(2x-y)

is 0 .

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