If sin A = -frac{60}{61},cot B = -frac{40}{9} | vedclass

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(C) Given $\sin A = -\frac{60}{61}$. Since $A$ is not in the $4^{	ext{th}}$ quadrant and $\sin A < 0$,$A$ must be in the $3^{	ext{rd}}$ quadrant. In

Vedclass · Accepted Answer

$-3$

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(C) Given $\sin A = -\frac{60}{61}$. Since $A$ is not in the $4^{	ext{th}}$ quadrant and $\sin A  0$. $\cos A = -\sqrt{1 - \sin^2 A} = -\sqrt{1 - (-\frac{60}{61})^2} = -\sqrt{\frac{3721 - 3600}{3721}} = -\frac{11}{61}$. Thus,$\cot A = \frac{\cos A}{\sin A} = \frac{-11/61}{-60/61} = \frac{11}{60}$.Given $\cot B = -\frac{40}{9}$. Since $B$ is not in the $4^{	ext{th}}$ quadrant and $\cot B Now,$6 \cot A + 4 \sec B = 6(\frac{11}{60}) + 4(-\frac{41}{40}) = \frac{11}{10} - \frac{41}{10} = -\frac{30}{10} = -3$.

If $\sin A = -\frac{60}{61}$,$\cot B = -\frac{40}{9}$ and neither $A$ nor $B$ is in the $4^{\text{th}}$ quadrant,then $6 \cot A + 4 \sec B = $

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