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If A lies in second quadrant 3tanA + 4 = 0, then the value of 2cotA − 5cosA + sinA is equal to

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Question

If A lies in second quadrant 3tan A + 4 = 0, then the value of 2cot A − 5cosA + sin A is equal to

Options

  • \[- \frac{53}{10}\]

     

  • \[\frac{23}{10}\]

     

  • \[\frac{37}{10}\]

     

  • \[\frac{7}{10}\]

     

MCQ
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Solution

It is given that \[\frac{\pi}{2} < A < \pi\].

\[3\tan A + 4 = 0\]

\[ \Rightarrow \tan A = - \frac{4}{3}\]

\[ \Rightarrow \cot A = - \frac{3}{4}\]
Now,
\[\sec A = \pm \sqrt{1 + \tan^2 A} = \pm \sqrt{1 + \frac{16}{9}} = \pm \sqrt{\frac{25}{9}} = \pm \frac{5}{3}\]
\[ \therefore \sec A = - \frac{5}{3} \left( \text{ A lies in 2nd quadrant }\right)\]
\[ \Rightarrow \cos A = - \frac{3}{5}\]
Also,

\[\sin A = \pm \sqrt{1 - \cos^2 A} = \pm \sqrt{1 - \frac{9}{25}} = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}\]

\[ \therefore \sin A = \frac{4}{5} \left( \text{ A lies in 2nd quadrant }\right)\]

So,
\[2\cot A - 5\cos A + \sin A\]
\[ = 2 \times \left( - \frac{3}{4} \right) - 5 \times \left( - \frac{3}{5} \right) + \frac{4}{5}\]
\[ = - \frac{3}{2} + 3 + \frac{4}{5}\]
\[ = \frac{- 15 + 30 + 8}{10}\]
\[ = \frac{23}{10}\]

Hence, the correct answer is option B.

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Chapter 5: Trigonometric Functions - Exercise 5.5 [Page 42]

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R.D. Sharma Mathematics [English] Class 11
Chapter 5 Trigonometric Functions
Exercise 5.5 | Q 20 | Page 42

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