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TS EAMCET 18-July-2022 Shift 1 Solved Paper

Section: Mathematics

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Question : 21 of 160

Marks: +1, -0

If

A

B(A > B)

are acute angles,

\sin (A - B) = \; \frac{16}{65}

\sin B = \; \frac{5}{13}

\tan A + \cot A =

$\frac{25}{12}$
$\frac{12}{25}$
$\frac{5}{12}$
$\frac{12}{5}$

Solution: 👈: Video Solution

A>B

and

A, B

are acute angles.

Given,

\sin (A-B) = \; \frac{16}{65}

and

\sin B = \; \frac{5}{13}

Then,

\cos (A-B) = \; \frac{63}{65}

and

\cos B = \; \frac{12}{13}

\sin A = \sin ((A-B)+B) = \sin (A-B) \cos B + \cos (A-B) \sin B

= \;\; \frac{16}{65} \times \frac{12}{13} + \frac{63}{65} \times \frac{5}{13} = \frac{507}{845} = \frac{3}{5}

\cos A = \cos ((A-B)+B) = \cos (A-B) \cos B - \sin (A-B) \sin B

= \;\; \frac{63}{65} \times \frac{12}{13} - \frac{16}{65} \times \frac{5}{13} = \frac{676}{845} = \frac{4}{5}

\therefore \;\; \tan A + \cot A = \frac{ \sin A }{ \cos A } + \frac{ \cos A }{ \sin A }

= \frac{1}{ \sin A \cos A } = \frac{1}{ \frac{3}{5} \times \frac{4}{5} } = \frac{25}{12}

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