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3.Trigonometrical Ratios, Functions and Identities
normal
જો $A$ એ તૃતીય ચરણમાં હોય અને $3\ tanA - 4 = 0$ થાય તો $5\ sin\ 2A + 3\ sinA + 4\ cosA$ =
A
$0$
B
$1$
C
$2$
D
none of these
Solution
$\tan A=\frac{4}{3}$
$\therefore \sin A=\frac{-4}{5}, \cos A=\frac{-3}{5}$
($\because$ A lies in third quadrant)
$5 \sin 2 A+3 \sin A+4 \cos A$
$=\frac{5.2 \tan A}{1-\tan ^{2} A}+3 \sin A+4 \cos A$
${=\frac{10 \cdot \frac{4}{3}}{1+\frac{16}{9}}-3 \cdot \frac{4}{5}-4 \cdot \frac{3}{5}}$
${=\frac{40}{3} \times \frac{9}{25}-\frac{12}{5}-\frac{12}{5}=\frac{24}{5}-\frac{24}{5}=0}$
Standard 11
Mathematics
