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UPSC NDA I 2014 Mathematics Paper

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Question : 50 of 120

Marks: +1, -0

What is

\;\frac{\cot 224^{\circ} - \cot 134^{\circ}}{\cot 226^{\circ} + \cot 316^{\circ}}

–cosec 88°
–cosec 2°
–cosec 44°
–cosec 46°

Solution:

We can write,

\cot (224^{\circ}) = \cot (44^{\circ})

\cot (134^{\circ}) = -\cot (46^{\circ})

\cot (226^{\circ}) = \cot (46^{\circ})

\cot (316^{\circ}) = -\cot (44^{\circ})

We can rewrite the equation as

\;\frac{\cot (224^{\circ}) - \cot (134^{\circ})}{\cot (226^{\circ}) + \cot (316^{\circ})}

\;\; = \;\frac{\cot (44^{\circ}) + \cot (46^{\circ})}{\cot (46^{\circ}) - \cot (44^{\circ})}

\;\text{ Now, }\; \cot A - \cot B = \;\frac{\cos A}{\sin A} - \;\frac{\cos B}{\sin B}

\cot A - \cot B = \;\frac{\sin (B-A)}{\sin A \sin B}

Similarly,

\cot A + \cot B = \;\frac{\sin (B+A)}{\sin A \sin B}

Putting

A = 46^{\circ}

and

B = 44^{\circ}

we get

\;\frac{\cot (44^{\circ}) + \cot (46^{\circ})}{\cot (46^{\circ}) - \cot (44^{\circ})}

\;\; = \;\frac{\sin (90^{\circ})}{\sin (-2^{\circ})}

\;\; = -\csc 2^{\circ}

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