Question : If $1+2 \tan ^2 \theta+2 \sin \theta \sec ^2 \theta=\frac{a}{b}, 0^{\circ}<\theta<90^{\circ}$, then $\frac{a+b}{a-b}=?$
Option 1: $\sin \theta$
Option 2: $\cos \theta$
Option 3: $\operatorname{cosec} \theta$
Option 4: $\sec \theta$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
Correct Answer: $\operatorname{cosec} \theta$
Solution :
According to the question
$1+2 \tan ^2 \theta+2 \sin \theta \sec ^2 \theta=\frac{a}{b}$
⇒ 1 + $\frac{2\sin^{2}\theta}{\cos^{2}\theta}$ + $\frac{2\sin^{2}\theta}{\cos^{2}\theta}=\frac{a}{b}$
⇒ $\frac{\cos^{2}\theta + 2\sin^{2}\theta + 2\sin\theta }{\cos^{2}\theta}= \frac{a}{b}$
⇒ $\frac{[1- \sin^{2}\theta + 2\sin^{2}\theta + 2\sin\theta]}{[1 - \sin^{2}\theta]}= \frac{a}{b}$
⇒ $\frac{[1+ \sin^{2}\theta + 2\sin\theta]}{[1 - \sin^{2}\theta]}= \frac{a}{b}$
⇒ $\frac{(1+ \sin\theta)^{2}}{(1- \sin\theta)(1+\sin\theta)}= \frac{a}{b}$
⇒ $\frac{1+ \sin\theta}{1- \sin\theta}= \frac{a}{b}$
So, $\frac{a + b}{a - b}$ = $\frac{1+ \sin\theta + 1- \sin\theta}{1+ \sin\theta - (1- \sin\theta)}$
= $\frac{2}{2\sin\theta}$
= $\operatorname{cosec}\theta$
Hence, the correct answer is $\operatorname{cosec} \theta$.
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Get Updates Brochure
Your Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.
