Question : If $\cot \theta=\frac{1}{\sqrt{3}}, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{2-\sin ^2 \theta}{1-\cos ^2 \theta}+\left(\operatorname{cosec}^2 \theta-\sec \theta\right)$ is:
Option 1: 0
Option 2: 2
Option 3: 5
Option 4: 1
Correct Answer: 1
Solution : $\cot \theta=\frac{1}{\sqrt{3}}$ ⇒ $\theta=60^{\circ}$ So, $\frac{2-\sin ^2 \theta}{1-\cos ^2 \theta}+\left(\operatorname{cosec}^2 \theta-\sec \theta\right)$ = $\frac{2-\sin ^2 60^{\circ}}{1-\cos ^2 60^{\circ}}+\left(\operatorname{cosec}^2 60^{\circ} -\sec 60^{\circ}\right)$ = $\frac{2- \frac{3}{4}}{1-\frac{1}{4}}+\frac{4}{3} -2$ = $\frac{\frac{5}{4}}{\frac{3}{4}}+\frac{4}{3} -2$ = $\frac{5}{3}+\frac{4}{3} -2$ = $\frac{9}{3} -2$ = 3 – 2 = 1 Hence, the correct answer is 1.
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Question : If $\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}=4,0^{\circ}<\theta<90^{\circ}$, then what is the value of $(\sec \theta+\operatorname{cosec} \theta+\cot \theta) ?$
Question : If $\frac{1}{\operatorname{cosec} \theta+1}+\frac{1}{\operatorname{cosec} \theta-1}=2 \sec \theta, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{\tan \theta+2 \sec \theta}{\operatorname{cosec} \theta}$ is:
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