Question : Which of the following is the value of $\sqrt{\frac{1-\sin 45^{\circ}}{1+\sin 45^{\circ}}}$?
Option 1: $\cos 45^{\circ} - \tan 45^{\circ}$
Option 2: $\tan 45^{\circ} - \sec 45^{\circ}$
Option 3: $\tan 45^{\circ}$
Option 4: $\sec 45^{\circ} - \tan 45^{\circ}$
Correct Answer: $\sec 45^{\circ} - \tan 45^{\circ}$
Solution : $\sqrt{\frac{1-\sin 45^{\circ}}{1+\sin 45^{\circ}}}$ = $\sqrt{\frac{(1-\sin 45^{\circ})^2}{(1+\sin 45^{\circ})(1-\sin 45^{\circ})}}$ = $\sqrt{\frac{(1-\sin 45^{\circ})^2}{(1-\sin^2 45^{\circ})}}$ = $\sqrt{\frac{(1-\sin 45^{\circ})^2}{\cos^2 45^{\circ}}}$ = $\frac{1-\sin 45^{\circ}}{\cos 45^{\circ}}$ = $\sec 45^{\circ} - \tan 45^{\circ}$ Hence, the correct answer is $\sec 45^{\circ} - \tan 45^{\circ}$.
Application | Eligibility | Selection Process | Result | Cutoff | Admit Card | Preparation Tips
Question : The value of $\frac{\sin ^2 30^{\circ}+\cos ^2 60^{\circ}+\sec 45^{\circ} × \sin 45^{\circ}}{\sec 60^{\circ}+{\text{cosec}} 30^{\circ}}$ is:
Option 1: $\frac{1}{4}$
Option 2: $-\frac{3}{8}$
Option 3: $\frac{3}{8}$
Option 4: $-\frac{1}{4}$
Question : Find the value of $\sqrt{\frac{1-\tan A}{1+\tan A}}$.
Option 1: $\sqrt{\frac{1+\sin 2 A}{\cos 2 A}}$
Option 2: $\sqrt{\frac{1-\sin 2 A}{\cos 2 A}}$
Option 3: $\sqrt{\frac{1+\sin A}{\cos A}}$
Option 4: $\sqrt{\frac{1-\sin A}{\cos A}}$
Question : If $k\left(\tan 45^{\circ} \sin 60^{\circ}\right)=\cos 60^{\circ} \cot 30^{\circ}$, then the value of k is:
Option 1: $1$
Option 2: $2$
Option 3: $\frac{1}{\sqrt{3}}$
Option 4: $\sqrt{3}$
Question : If $4-2 \sin ^2 \theta-5 \cos \theta=0,0^{\circ}<\theta<90^{\circ}$, then the value of $\cos \theta-\tan \theta$ is:
Option 1: $\frac{1+2 \sqrt{3}}{2}$
Option 2: $\frac{2-\sqrt{3}}{2}$
Option 3: $\frac{2+\sqrt{3}}{2}$
Option 4: $\frac{1-2 \sqrt{3}}{2}$
Question : If $4-2 \sin ^2 \theta-5 \cos \theta=0,0^{\circ}<\theta<90^{\circ}$, then the value of $\cos \theta+\tan \theta$ is:
Option 1: $\frac{2-\sqrt{3}}{2}$
Option 2: $\frac{1+2 \sqrt{3}}{2}$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile