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: $\sqrt{\frac{1-\sin A}{\cos A}}$
Correct Answer: $\sqrt{\frac{1-\sin 2 A}{\cos 2 A}}$
Solution : Given expression, $\sqrt{\frac{1-\tan A}{1+\tan A}}$ = $\sqrt{\frac{1-\frac{\sin A}{\cos A}}{1+\frac{\sin A}{\cos A}}}$ = $\sqrt{\frac{\cos A-\sin A}{\cos A+\sin A}}$ Multiplying and dividing by $\sqrt{\cos A-\sin A}$ $\sqrt{\frac{\cos A-\sin A}{\cos A+\sin A}}\times\sqrt{\frac{\cos A-\sin A}{\cos A-\sin }}$ = $\sqrt{\frac{(\cos A-\sin A)^2}{(\cos A+\sin A)(\cos A-\sin A)}}$ = $\sqrt{\frac{\sin^2A+\cos^2A-2\sin A \cos A}{\cos^2 A-\sin^2 A}}$ We know, $\sin^2\theta+\cos^2\theta=1,\ \sin2\theta=2\sin\theta\cos\theta\text{ and }\cos2\theta=\cos^2 \theta-\sin^2 \theta$ = $\sqrt{\frac{1-\sin2A}{\cos2A}}$ Hence, the correct answer is $\sqrt{\frac{1-\sin2A}{\cos2A}}$.
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Question : If $\operatorname{cos} \theta+\operatorname{sin} \theta=\sqrt{2} \operatorname{cos} \theta$, find the value of $(\cos \theta-\operatorname{sin} \theta)$
Option 1: $\sqrt{2} \sin \theta$
Option 2: $\sqrt{2} \cos \theta$
Option 3: $\frac{1}{\sqrt{2}} \sin \theta$
Option 4: $\frac{1}{2}\cos \theta$
Question : If $\sin A=\frac{1}{2}$, then the value of $(\tan A+\cos A)$ is:
Option 1: $\frac{2}{3 \sqrt{3}}$
Option 2: $\frac{3}{2 \sqrt{3}}$
Option 3: $\frac{5}{2 \sqrt{3}}$
Option 4: $\frac{5}{3 \sqrt{3}}$
Question : If $\sin (A-B)=\sin A \cos B–\cos A\sin B$, then $\sin 15°$ will be:
Option 1: $\frac{\sqrt{3}+1}{2\sqrt{2}}$
Option 2: $\frac{\sqrt{3}}{2\sqrt{2}}$
Option 3: $\frac{\sqrt{3}–1}{–\sqrt{2}}$
Option 4: $\frac{\sqrt{3}–1}{2\sqrt{2}}$
Question : If $\sin A=\frac{2}{3}$, then find the value of (7 – tan A)(3 + cos A).
Option 1: $\frac{61}{3}+\frac{17}{3 \sqrt{5}}$
Option 2: $\frac{61}{3 \sqrt{5}}+\frac{17}{3}$
Option 3: $\frac{61}{3}+\frac{17}{\sqrt{5}}$
Option 4: $\frac{61}{3}-\frac{17}{3 \sqrt{5}}$
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: $\frac{1-2 \sqrt{3}}{2}$
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