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: $\sec \theta$
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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 $\operatorname{cosec} \theta$.
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Question : $\left(\frac{\tan ^3 \theta}{\sec ^2 \theta}+\frac{\cot ^3 \theta}{\operatorname{cosec}^2 \theta}+2 \sin \theta \cos \theta\right) \div\left(1+\operatorname{cosec}^2 \theta+\tan ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : $\frac{(1+\sec \theta \operatorname{cosec} \theta)^2(\sec \theta-\tan \theta)^2(1+\sin \theta)}{(\sin \theta+\sec \theta)^2+(\cos \theta+\operatorname{cosec} \theta)^2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : $\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)} \times \frac{\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}}{\tan \theta+\cot \theta}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : Let $0^{\circ}<\theta<90^{\circ}$, $\left(1+\cot ^2 \theta\right)\left(1+\tan ^2 \theta\right) × (\sin \theta-\operatorname{cosec} \theta)(\cos \theta-\sec \theta)$ is equal to:
Question : The expression $\frac{(1-\sin \theta+\cos \theta)^2(1-\cos \theta) \sec ^3 \theta\; {\operatorname{cosec}}^2 \theta}{(\sec \theta-\tan \theta)(\tan \theta+\cot \theta)}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
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