Question

If $3\sin \theta +4\cos \theta =5$, then find $\cot \theta +\text{cosec}\theta$

• A. 1
• B. -1
• C. 3
• D. 2

Answer 1

Answer: (C)

3

Given, $3\sin \theta +4\cos \theta =5$
$3\sin \theta =5-4\cos \theta$, squaring we get:
$9{\sin }^2\theta =25+16{\cos }^2\theta -40\cos \theta$
$9(1-{\cos }^2\theta )=25+16{\cos }^2\theta -40\cos \theta (\because {\sin }^2\theta =1-{\cos }^2\theta )$
$9-9{\cos }^2\theta =25+16{\cos }^2\theta -40\cos \theta$
$\Rightarrow 16{\cos }^2\theta +9{\cos }^2\theta -40\cos \theta +25-9=0$
$\Rightarrow 25{\cos }^2\theta -40\cos \theta +16=0$
$\Rightarrow (5\cos \theta -4{)}^2=0$
$\Rightarrow 5\cos \theta -4=0$
$\Rightarrow 5\cos \theta =4$
$\Rightarrow \cos \theta =\frac{4}{5}$
$\text{cosec}\theta =\frac{H}{P}=\frac{OP}{PQ}=\frac{5}{3}$
$\cot \theta =\frac{B}{P}=\frac{OQ}{PQ}=\frac{4}{3}$
Required Expression $=\text{cosec}\theta +\cot \theta$
$=\frac{5}{3}+\frac{4}{3}=\frac{4}{3}=\frac{9}{3}=3$