If A lies in the second quadrant and 3tan A + | vedclass

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Question

(d) $3\,	an A + 4 = 0\, \Rightarrow \,	an A = - \frac{4}{3}$ 

$ \Rightarrow \,\,\sin A\, = \pm \,\frac{{	an A}}{{\sqrt {1 + {{	an }^2}A} }

Vedclass · Accepted Answer

$\frac{{23}}{{10}}$

Vedclass · Answer

(d) $3\,	an A + 4 = 0\, \Rightarrow \,	an A = - \frac{4}{3}$ 

$ \Rightarrow \,\,\sin A\, = \pm \,\frac{{	an A}}{{\sqrt {1 + {{	an }^2}A} }} $

$= \pm \frac{{ - 4/3}}{{\sqrt {1 + 16/9} }} = \frac{4}{5}$        ($\because$ $A$ is in $2^{nd}$ quadrant)

and $\cos \,A = - \frac{3}{5}$. 

Thus, $2\cot A - 5\cos A + \sin A$ $ = 2\,\left( { - \frac{3}{4}} ight) - 5\,\left( { - \frac{3}{5}} ight) + \frac{4}{5} = \frac{{23}}{{10}}$.

If $A$ lies in the second quadrant and $3\tan A + 4 = 0,$ the value of $2\cot A - 5\cos A + \sin A$ is equal to

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