Question : If $4\sin^{2}\theta-1=0$ and angle $\theta$ is less then $90^{\circ}$, the value of $\cos^{2}\theta+\tan^{2}\theta$ is: (Take $0^{\circ}< \theta< 90^{\circ}$)
Option 1: $\frac{17}{15}$
Option 2: $\frac{13}{12}$
Option 3: $\frac{11}{9}$
Option 4: $\frac{12}{11}$
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Correct Answer: $\frac{13}{12}$
Solution : Given: $4\sin^{2}\theta-1=0$ $⇒4\sin^{2}\theta-1=0$ $⇒4\sin^{2}\theta=1$ $⇒\sin^{2}\theta=\frac{1}{4}$ $⇒\sin\theta=\pm\frac{1}{2}$ Since we're given that $0^{\circ}< \theta< 90^{\circ}$, So, $\sin\theta$ is positive in this range. $⇒\sin\theta=\frac{1}{2}=\sin30^\circ$ $\therefore \theta = 30^\circ$ Putting the value of $\theta$, we get, $\therefore\cos^{2}\theta+\tan^{2}\theta=\cos^{2}30^\circ+\tan^{2}30^\circ=(\frac{\sqrt3}{2})^2+(\frac{1}{\sqrt3})^2=\frac{3}{4}+\frac{1}{3}=\frac{9+4}{12}=\frac{13}{12}$ Hence, the correct answer is $\frac{13}{12}$.
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Question : If $\sin (\theta +18^{\circ})=\cos 60^{\circ}(0< \theta < 90^{\circ})$, then the value of $\cos 5\theta$ is:
Option 1: $\frac{1}{2}$
Option 2: $0$
Option 3: $\frac{1}{\sqrt{2}}$
Option 4: $1$
Question : If $\theta$ is an acute angle and $\cos\theta=\frac{11}{17}$, what is the value of $\tan\theta$?
Option 1: $\frac{4 \sqrt{10}}{11}$
Option 2: $\frac{13}{11}$
Option 3: $\frac{2 \sqrt{42}}{11}$
Option 4: $\frac{2 \sqrt{42}}{17}$
Question : If $\frac{\cos \theta}{(1+\sin \theta)}+\frac{\cos \theta}{(1-\sin \theta)}=4$ and $\theta$ is acute, then the value of $\theta$ is:
Option 1: $60^{\circ}$
Option 2: $15^{\circ}$
Option 3: $45^{\circ}$
Option 4: $30^{\circ}$
Question : If $\theta$ is a positive acute angle and $4\cos ^{2}\theta -4\cos \theta +1=0$, then the value of $\tan (\theta -15^{\circ})$is equal to:
Option 1: $0$
Option 2: $1$
Option 3: $\sqrt{3}$
Option 4: $\frac{1}{\sqrt{3}}$
Question : If $\theta$ is an acute angle and $\sin \theta \cos \theta=2 \cos ^3 \theta-\frac{1}{4} \cos \theta$, then the value of $\sin \theta$ is:
Option 1: $\frac{\sqrt{15}-1}{8}$
Option 2: $\frac{\sqrt{15}-1}{4}$
Option 3: $\frac{\sqrt{15}+1}{4}$
Option 4: $\frac{\sqrt{15}-1}{2}$
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