Question : Which of the following is equal to $[\frac{\tan \theta+\sec \theta–1}{\tan \theta–\sec \theta+1}]$?
Option 1: $\frac{1+\sin \theta}{\cos \theta}$
Option 2: $\frac{1+\tan \theta}{\cot \theta}$
Option 3: $\frac{1+\cot \theta}{\tan \theta}$
Option 4: $\frac{1+\cos \theta}{\sin \theta}$
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Correct Answer: $\frac{1+\sin \theta}{\cos \theta}$
Solution : Given: The given trigonometric expression is $[\frac{\tan \theta+\sec \theta–1}{\tan \theta–\sec \theta+1}]$. Use the trigonometric identity, $\sec^2\theta–\tan^2\theta=1$ ⇒ $[\frac{\tan \theta+\sec \theta–(\sec^2\theta–\tan^2\theta)}{\tan \theta–\sec \theta+1}]=[\frac{(\tan \theta+\sec \theta)(1–\sec\theta+\tan \theta)}{\tan \theta–\sec \theta+1}$ $=\tan \theta+\sec \theta=\frac{1+\sin \theta}{\cos \theta}$ Hence, the correct answer is $\frac{1+\sin \theta}{\cos \theta}$.
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Question : Which of the following is equal to $[\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}]$?
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:
Question : $\frac{1+\sin \theta}{\cos \theta}$ is equal to which of the following (where $\left.\theta \neq \frac{\pi}{2}\right)?$
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 : $\left(\frac{\tan ^3 \theta}{\sec ^2 \theta}+\frac{\cot ^3 \theta}{\operatorname{cosec}^2 \theta}+2 \sin \theta \cos \theta\right) \div\left(1+\operatorname{cosec}^2 \theta+\tan ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, is equal to:
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