Test Index

TG EAMCET 2 May 2025 Shift 1 Paper

© examsnet.com

Question : 22 of 160

Marks: +1, -0

If

\sin A = -\frac{60}{61}, \cot B = -\frac{40}{9}

A

B

is in 4th quadrant, then

6 \cot A + 4 \sec B =

$rac{26}{5}$
$-\frac{26}{5}$
-3
3

Solution:

Given,

\sin A = -\frac{60}{61}, \cot B = -\frac{40}{9}

Now,

\cos^2 A = 1 - \sin^2 A

\Rightarrow 1 - \left(-\frac{60}{61}\right)^2 \Rightarrow 1 - \frac{3600}{3721} = \frac{121}{3721}

\therefore \cos A = \pm \frac{11}{61}

But

\sin A < 0

and

A

and

B

are not in 4th quadrant.

So,

A

in third quadrant and

B

in 2nd quadrant

\therefore \cos A = -\frac{11}{61}

And

\cot A = \frac{\cos A}{\sin A} = \frac{-11}{-60} = \frac{11}{60}

6 \cot A = \frac{66}{60} = \frac{11}{10}

Now,

\tan B = \frac{1}{\cot B} = \frac{-9}{40}

And,

\sec^2 B = 1 + an^2 B

\Rightarrow 1 + \left(\frac{-9}{40}\right)^2 = 1 + \frac{81}{1600} = \frac{41}{40}

\Rightarrow \sec B = -\frac{41}{40} (\because B \text{ in 2nd quadrant})

\Rightarrow 4 \sec B = \frac{-164}{40} = -\frac{41}{40}

\therefore 6 \cot A + 4 \sec B = \frac{11}{10} - \frac{41}{10}

\Rightarrow \frac{-30}{10} = -3

© examsnet.com

Go to Question:

Prev Question

Next Question