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: $\frac{1-2 \sqrt{3}}{2}$
Correct Answer: $\frac{1-2 \sqrt{3}}{2}$
Solution : Given: $4 - 2\sin^2 \theta - 5\cos \theta = 0$ ⇒ $4 - 2 × (1 - \cos^2\theta) - 5cos\theta = 0$ ⇒ $4 - 2 + 2\cos^2\theta - 5\cos\theta = 0$ ⇒ $2\cos^2\theta - 5\cos\theta + 2 = 0$ ⇒ $2\cos^2\theta - 4\cos\theta - \cos\theta + 2 = 0$ ⇒ $2\cos\theta × (\cos\theta - 2) - 1 × (\cos\theta - 2) = 0$ ⇒ $(2\cos\theta - 1) × (\cos\theta - 2) = 0$ ⇒ $\cos\theta = \frac{1}{2}, 2$ Rejecting $\cos\theta = 2$ as $0^\circ < \theta < 90^\circ$ So, $\cos\theta=\frac{1}{2}$ ⇒ $\theta = 60^\circ$ The value of $\cos \theta -\tan \theta = \cos60^\circ - \tan60^\circ$ = $\frac{1}{2} -\sqrt3$ = $\frac{1-2 \sqrt{3}}{2}$ Hence, the correct answer is $\frac{1-2 \sqrt{3}}{2}$.
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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: $\frac{1+2 \sqrt{3}}{2}$
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 $\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}=4,0^{\circ}<\theta<90^{\circ}$, then what is the value of $(\sec \theta+\operatorname{cosec} \theta+\cot \theta) ?$
Option 1: $1+2 \sqrt{3}$
Option 2: $\frac{1+2 \sqrt{3}}{3}$
Option 3: $\frac{2+\sqrt{3}}{3}$
Option 4: $2+\sqrt{3}$
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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