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Trigonometric Functions — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions1 Mark
MCQ
If $A$ lies in second quadrant $3\tan\text{A}+4=0,$ then the value of $2\cot\text{A}-5\cot\text{A}+\sin\text{A}$ is :
  • A
    $-\frac{53}{10}$
  • ✓
    $\frac{23}{10}$
  • C
    $\frac{37}{10}$
  • D
    $\frac{7}{10}$
✓

Answer

Correct option: B.
$\frac{23}{10}$
It is given that $\frac{\pi}{2}<\text{A}<\pi$
$3\tan\text{A}+4=0$
$\Rightarrow\tan\text{A}=-\frac{4}{3}$
$\Rightarrow\cot\text{A}=-\frac{3}{4}$
Now,
$\sec\text{A}=\pm\sqrt{1+\tan^2\text{A}}$
$=\pm\sqrt{1+\frac{16}{9}}$
$=\pm\sqrt{\frac{25}{9}}=\pm\frac{5}{3}$
Also,
$\sin\text{A}=\pm\sqrt{1-\cos^2\text{A}}$
$=\pm\sqrt{1-\frac{9}{25}}$
$=\pm\sqrt{\frac{16}{25}}=\pm\frac{4}{5}$
$\therefore\sin\text{A}=\frac{4}{5} \ ($A lines in $2$ nd quadrant$)$
so $, 2\cot\text{A}-5\cos\text{A}+\sin\text{A}$
$=2\times\Big(-\frac{3}{5}\Big)-5\times\Big(-\frac{3}{5}\Big)+\frac{4}{5}$
$=-\frac{3}{2}+3+\frac{4}{5}$
$=-\frac{15+30+8}{10} $
$=\frac{23}{10}$
Hence,the correct answer is option $B$.

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