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3.Trigonometrical Ratios, Functions and Identities
easy
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
A
$\frac{{ - 53}}{{10}}$
B
$\frac{{ - 7}}{{10}}$
C
$\frac{7}{{10}}$
D
$\frac{{23}}{{10}}$
Solution
(d) $3\,\tan A + 4 = 0\, \Rightarrow \,\tan A = – \frac{4}{3}$
$ \Rightarrow \,\,\sin A\, = \pm \,\frac{{\tan A}}{{\sqrt {1 + {{\tan }^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}} \right) – 5\,\left( { – \frac{3}{5}} \right) + \frac{4}{5} = \frac{{23}}{{10}}$.
Standard 11
Mathematics
