Question : Simplify $\frac{1 + \sin t}{4 - 4 \sin t} - \frac{1 - \sin t}{4 + 4 \sin t}$.
Option 1: $4\tan t .\sin t$
Option 2: $\tan t . \sec t$
Option 3: $\tan t - \sin t$
Option 4: $\tan t + \sin t$
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Correct Answer: $\tan t . \sec t$
Solution : Given: $\frac{1 + \sin t}{4 -4\sin t} - \frac{1 - \sin t}{4 + 4\sin t}$ $=\frac{1 + \sin t}{4(1 - \sin t)} - \frac{1 - \sin t}{4(1 + \sin t)}$ $=\frac{(1 + \sin t)^{2} - (1 - \sin t)^{2}}{4 × (1+ \sin t)(1 - \sin t)}$ $=\frac{4\sin t}{4(1 - \sin^{2} t)}$ $=\frac{\sin t}{\cos^{2} t}$ $=\frac{\sin t}{\cos t} × \frac{1}{\cos t}$ $=\tan t.\sec t$ Hence, the correct answer is $\tan t.\sec t$.
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Question : Find the value of: $\sqrt{\frac{1 - \sin 3 \theta}{1 + \sin 3 \theta}}$
Option 1: $\sec 3 \theta - \tan 3 \theta$
Option 2: $(\sec 3 \theta - \tan 3 \theta)^3$
Option 3: $(\sec 3 \theta - \tan 3 \theta)^2$
Option 4: $\sec 3 \theta + \tan 3 \theta$
Question : Simplify the following. $\frac{\sin^3 \alpha+\cos^3 \alpha}{\sin \alpha+\cos \alpha}$
Option 1: $1+\sin \alpha \cos \alpha$
Option 2: $\tan \alpha$
Option 3: $1-\sin \alpha \cos \alpha$
Option 4: $\sec \alpha$
Question : If $\sec A=\frac{5}{4}$, then the value of $\frac{\tan A}{1+\tan ^2 A}-\frac{\sin A}{\sec A}$ is:
Option 1: 2
Option 2: 1
Option 3: 0
Option 4: 3
Question : Simplify the following expression. $\frac{\sin \theta - 2 \sin ^3 \theta}{2 \cos ^3 \theta - \cos \theta}$
Option 1: $\tan \theta$
Option 2: $\sin \theta$
Option 3: $\sec \theta$
Option 4: $\cos \theta$
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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