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3.Trigonometrical Ratios, Functions and Identities
medium
જો $A$ એ ત્રીજા ચરણમાં હોય અને $3\,\tan A - 4 = 0,$ તો $5\,\sin 2A + 3\,\sin A + 4\,\cos A = $
A
$0$
B
$\frac{{ - 24}}{5}$
C
$\frac{{24}}{5}$
D
$\frac{{48}}{5}$
Solution
(a) $3\tan A – 4 = 0 \Rightarrow \tan A = \frac{4}{3} $
$\Rightarrow \sin A = – \frac{4}{5},\cos A = – \frac{3}{5}$
$\therefore $ $5\sin 2A + 3\sin A + 4\cos A$
$= 10\sin A\cos A + 3\sin A + 4\cos A$
$= 10\,\left( {\frac{{12}}{{25}}} \right) – \frac{{12}}{5} – \frac{{12}}{5} = 0$.
Standard 11
Mathematics
