Question : If $0 \leq \theta \leq 90^{\circ}$, and $\sin \left(2 \theta+50^{\circ}\right)=\cos \left(4 \theta+16^{\circ}\right)$, then what is the value of $\theta$ (in degrees)?
Option 1: $10^{\circ}$
Option 2: $8^{\circ}$
Option 3: $4^{\circ}$
Option 4: $12^{\circ}$
Correct Answer: $4^{\circ}$
Solution : Given, $\sin \left(2 \theta+50°\right)=\cos \left(4 \theta+16°\right)$ We know, $\sin(90°-\theta)=\cos\theta$ So, $\sin \left(2 \theta+50°\right)=\sin(90°-(4 \theta+16°)$ ⇒ $2\theta+50°=90°-4\theta-16°$ ⇒ $2\theta+4\theta=90°-16°-50°$ ⇒ $6\theta=24°$ $\therefore\theta=4°$ Hence, the correct answer is $4°$.
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Question : If $3\left(\cot ^2 \theta-\cos ^2 \theta\right)=1-\sin ^2 \theta, 0^{\circ}<\theta<90^{\circ}$, then $\theta$ is equal to:
Option 1: $30^{\circ}$
Option 2: $60^{\circ}$
Option 3: $45^{\circ}$
Option 4: $15^{\circ}$
Question : If $\cot \theta=\frac{1}{\sqrt{3}}, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{2-\sin ^2 \theta}{1-\cos ^2 \theta}+\left(\operatorname{cosec}^2 \theta-\sec \theta\right)$ is:
Option 1: 0
Option 2: 2
Option 3: 5
Option 4: 1
Question : If $3+\cos ^2 \theta=3\left(\cot ^2 \theta+\sin ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, then what is the value of $(\cos \theta+2 \sin \theta)$ ?
Option 1: $\frac{2 \sqrt{3}+1}{2}$
Option 2: $3 \sqrt{2}$
Option 3: $\frac{3 \sqrt{3}+1}{2}$
Option 4: $\frac{\sqrt{3}+2}{2}$
Question : If $(\cos \theta+\sin \theta):(\cos \theta-\sin \theta)=(\sqrt{3}+1):(\sqrt{3}-1), 0^{\circ}<\theta<90^{\circ}$, then what is the value of $\sec \theta$?
Option 1: 2
Option 2: $\sqrt{2}$
Option 3: 1
Option 4: $\frac{2 \sqrt{3}}{3}$
Question : Find the value of $\frac{\sin ^2 39^{\circ}+\sin ^2\left(90^{\circ}–39^{\circ}\right)}{\cos ^2 35^{\circ}+\cos ^2\left(90^{\circ}–35^{\circ}\right)}+3 \tan 25^{\circ} \tan 75^{\circ}$:
Option 2: 4
Option 3: 3
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