Question : Solve for $\theta: \cos ^2 \theta-\sin ^2 \theta=\frac{1}{2}, 0<\theta<90^{\circ}$.
Option 1: 45o
Option 2: 60o
Option 3: 30o
Option 4: 40o
Correct Answer: 30o
Solution : Given, cos2θ − sin2θ = $\frac{1}{2}$ ⇒ cos2θ = $\frac{1}{2}$ + sin2θ since, cos2θ + sin2θ = 1 ⇒ 1 − sin2θ = $\frac{1}{2}$ + sin2θ ⇒ sin2θ + sin2θ = $\frac{1}{2}$ ⇒ 2sin2θ = $\frac{1}{2}$ ⇒ sin2θ = $\frac{1}{4}$ ⇒ sin θ = ± $\frac{1}{2}$ Since, 0 < θ < 90º, sin θ = $\frac{1}{2}$ ⇒ θ = 30º Hence, the correct answer is 30º.
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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 $\frac{\cos ^2 \theta}{\cot ^2 \theta–\cos ^2 \theta}=3$, where $0^{\circ}<\theta<90^{\circ}$ then the value of $\theta$ is:
Option 1: $45^{\circ}$
Option 2: $50^{\circ}$
Option 3: $60^{\circ}$
Option 4: $30^{\circ}$
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 : 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 $\cot \theta=\frac{1}{\sqrt{3}}, 0^{\circ}<\theta<90^{\circ}$, 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
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