Question : $\theta$ is a positive acute angle and $\sin\theta-\cos\theta=0$, then the value of $\sec\theta+\operatorname{cosec}\theta$ is:
Option 1: $2$
Option 2: $\sqrt{3}$
Option 3: $2\sqrt{2}$
Option 4: $3\sqrt{2}$
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Correct Answer: $2\sqrt{2}$
Solution : Given: $\sin\theta-\cos\theta=0$ ⇒ $\sin\theta=\cos\theta$ ⇒ $\sin\theta=\sin\;(90°-\theta)$ ⇒ $2\theta=90°$ ⇒ $\theta=45°$ So, $\sec\theta+\operatorname{cosec}\theta$ $= \sec 45°+ \operatorname{cosec} 45°$ $= \sqrt{2}+\sqrt{2}$ $= 2\sqrt{2}$ Hence, the correct answer is $2\sqrt{2}$.
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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 : $\left(\frac{\tan ^3 \theta}{\sec ^2 \theta}+\frac{\cot ^3 \theta}{\operatorname{cosec}^2 \theta}+2 \sin \theta \cos \theta\right) \div\left(1+\operatorname{cosec}^2 \theta+\tan ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : $\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)} \times \frac{\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}}{\tan \theta+\cot \theta}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : $\frac{(1+\sec \theta \operatorname{cosec} \theta)^2(\sec \theta-\tan \theta)^2(1+\sin \theta)}{(\sin \theta+\sec \theta)^2+(\cos \theta+\operatorname{cosec} \theta)^2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : If $\cos \theta+\cos ^2 \theta=1$, find the value of $\sqrt{\sin ^4 \theta+\cos ^2 \theta}$.
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