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AP EAPCET 07-Jul-2022 Shift 1 Paper

Section: Mathematics

Question : 21 of 160

Marks: +1, -0

n

\; \frac{1}{\sin 45^{\circ} \sin 46^{\circ}} +\; \frac{1}{\sin 47^{\circ} \sin 48^{\circ}} + . . . +\; \frac{1}{\sin 133^{\circ} \sin 134^{\circ}} =\; \frac{1}{\sin (n^{\circ})}

Solution: 👈: Video Solution

\frac{1}{\sin 45^{\circ} \cdot \sin 46^{\circ}} + \frac{1}{\sin 47^{\circ} \cdot \sin 48^{\circ}} + . . . +

\frac{1}{\sin 133^{\circ} \cdot \sin 134^{\circ}}

= \frac{1}{\sin 1^{\circ}} \left[ \frac{\sin (46^{\circ}-45^{\circ})}{\sin 45^{\circ} \cdot \sin 46^{\circ}} + \frac{\sin (48^{\circ}-47^{\circ})}{\sin 47^{\circ} \cdot \sin 48^{\circ}} + . . . + .. \right.

\left. \frac{\sin (134^{\circ}-133^{\circ})}{\sin 133^{\circ} \cdot \sin 134^{\circ}} \right]

= \frac{1}{\sin 1^{\circ}} \left[ (\cot 45^{\circ}-\cot 46^{\circ}) + (\cot 47^{\circ}-\cot 48^{\circ}) . \right.

\left. + . . . + (\cot 133^{\circ}-\cot 134^{\circ}) \right]

= \frac{1}{\sin 1^{\circ}} \left[ (\cot 45^{\circ}-\cot 46^{\circ}) + (\cot 47^{\circ}-\cot 48^{\circ}) . \right.

\left. + . . . + (-\cot 47^{\circ}+\cot 46^{\circ}) \right]

\left[ \because \cot 133^{\circ}=\cot (180^{\circ}-47^{\circ}) =-\cot 47^{\circ} \right.

and

\left. \cot 134^{\circ}=\cot (180^{\circ}-46^{\circ}) =-\cot 46^{\circ} \right.

\left. \text{and so on} \right]

= \frac{1}{\sin 1^{\circ}} \times \cot 45^{\circ} = \frac{1}{\sin 1^{\circ}}

According to question,

\frac{1}{\sin (n^{\circ})} = \frac{1}{\sin 1^{\circ}}

\therefore n=1

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