Question : If $\frac{\tan\theta +\cot\theta }{\tan\theta -\cot\theta }=2, (0\leq \theta \leq 90^{0})$, then the value of $\sin\theta$ is:
Option 1: $\frac{2}{\sqrt3}$
Option 2: $\frac{\sqrt3}{2}$
Option 3: $\frac{1}{2}$
Option 4: $1$
Correct Answer: $\frac{\sqrt3}{2}$
Solution : $\frac{\tan\theta +\cot\theta }{\tan\theta -\cot\theta }=2$ Use rule of componendo and dividendo i.e. If $\frac{a}{b}=\frac{c}{d}$ then, $\frac{a+b}{a-b}=\frac{c+d}{c-d}$. we have, ⇒ $\frac{\tan\theta +\cot\theta +\tan\theta -\cot\theta}{\tan\theta +\cot\theta -\tan\theta +\cot\theta}=\frac{2+1}{2-1}$ ⇒$\frac{2\tan\theta}{2\cot\theta} = 3$ ⇒ $\tan^2\theta=3$ ⇒ $\sin^2\theta= 3\cos^2\theta$ ⇒ $\sin^2\theta= 3(1-\sin^2\theta)$ ⇒ $4\sin^2\theta= 3$ ⇒ $\sin\theta=\frac{ \sqrt3 }{2}$ Hence, the correct answer is $\frac{ \sqrt3 }{2}$.
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Question : If $1 + \sin^2 θ - 3\sinθ \cosθ = 0$, then the value of $\cotθ$ is:
Question : If $\tan A=\frac{4}{3}, 0 \leq A \leq 90^{\circ}$, then find the value of $\sin A$.
Question : Simplify $\frac{\cos ^4 \theta-\sin ^4 \theta}{\sin ^2 \theta}$.
Question : The value of $\frac{2 \cos ^3 \theta-\cos \theta}{\sin \theta-2 \sin ^3 \theta}$ is:
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:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile