Question : If $\cot^2θ = 1 - e^2$, then the value of $\operatorname{cosec} θ + \cot^3θ \sec θ$ is:
Option 1: $\left(2-{e}^2\right)^ \frac{1}{2}$
Option 2: $\left(1-{e}^2\right)^ \frac{3}{2}$
Option 3: $\left(1-{e}^2\right)$
Option 4: $\left(2-{e}^2\right) ^\frac{3}{2}$
Correct Answer: $\left(2-{e}^2\right) ^\frac{3}{2}$
Solution : Given, $\cot^2θ = 1 - e^2$ Consider, $\operatorname{cosec} θ + \cot^3θ \sec θ$ $=\frac{1}{\sinθ} + \frac{\cos^3θ}{\sin^3θ}\frac1{\cosθ}$ $=\frac{\sin^2θ+\cos^2θ}{\sin^3θ}$ $=\frac{1}{\sin^3θ}$ [As $\sin^2θ+\cos^2θ=1$] $=\operatorname{cosec^3}θ$ Also, we know that, $\operatorname{cosec^2}θ=1+\cot^2θ$ ⇒ $\operatorname{cosec^2}θ=1+1-e^2$ ⇒ $\operatorname{cosec^2}θ=2-e^2$ ⇒ $\operatorname{cosec}θ=(2-e^2)^{\frac12}$ ⇒ $\operatorname{cosec^3}θ=(2-e^2)^{\frac32}$ Hence, the correct answer is $(2-e^2)^{\frac32}$.
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Question : If $\frac{1}{\operatorname{cosec} \theta+1}+\frac{1}{\operatorname{cosec} \theta-1}=2 \sec \theta, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{\tan \theta+2 \sec \theta}{\operatorname{cosec} \theta}$ is:
Option 1: $\frac{4+\sqrt{2}}{2}$
Option 2: $\frac{2+\sqrt{3}}{2}$
Option 3: $\frac{4+\sqrt{3}}{2}$
Option 4: $\frac{2+\sqrt{2}}{2}$
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
Question : The value of $\tan ^2 48^{\circ}-\operatorname{cosec}^2 42^{\circ}+\operatorname{cosec}\left(67^{\circ}+\theta\right)-\sec \left(23^{\circ}-\theta\right)$ is:
Option 1: $-1$
Option 2: $0$
Option 3: $1$
Option 4: $-2$
Question : If $(\frac{\sec \theta-1}{\sec \theta+1})^n=\operatorname{cosec} \theta-\cot \theta$, then $n=?$
Option 1: 1
Option 2: 0.5
Option 3: –1
Option 4: –0.5
Question : If $\tan (11 \theta)=\cot (7 \theta)$, then what is the value of $\sin ^2(6 \theta)+\sec ^2(9 \theta)+\operatorname{cosec}^2(12 \theta) ?$
Option 1: $\frac{23}{6}$
Option 2: $\frac{35}{12}$
Option 3: $\frac{31}{12}$
Option 4: $\frac{43}{12}$
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