If tan theta = frac{1}{sqrt{11}} and 0

Question

(C) Given $	an 	heta = \frac{1}{\sqrt{11}}.$We know that $\sec^{2} 	heta = 1 + 	an^{2} 	heta = 1 + \frac{1}{11} = \frac{12}{11}.$Also,$\operatorn

Vedclass · Accepted Answer

$\frac{5}{6}$

Vedclass · Answer

(C) Given $	an 	heta = \frac{1}{\sqrt{11}}.$We know that $\sec^{2} 	heta = 1 + 	an^{2} 	heta = 1 + \frac{1}{11} = \frac{12}{11}.$Also,$\operatorname{cosec}^{2} 	heta = 1 + \cot^{2} 	heta = 1 + (\frac{1}{	an 	heta})^{2} = 1 + (\sqrt{11})^{2} = 1 + 11 = 12.$Now,substitute these values into the expression:$\frac{\operatorname{cosec}^{2} 	heta - \sec ^{2} 	heta}{\operatorname{cosec}^{2} 	heta + \sec ^{2} 	heta} = \frac{12 - \frac{12}{11}}{12 + \frac{12}{11}}.$Divide the numerator and denominator by $12$:$\frac{1 - \frac{1}{11}}{1 + \frac{1}{11}} = \frac{\frac{10}{11}}{\frac{12}{11}} = \frac{10}{12} = \frac{5}{6}.$

If $\tan \theta = \frac{1}{\sqrt{11}}$ and $0 < \theta < \frac{\pi}{2},$ then the value of $\frac{\operatorname{cosec}^{2} \theta - \sec ^{2} \theta}{\operatorname{cosec}^{2} \theta + \sec ^{2} \theta}$ is

Similar Questions

The value of $\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}$ is

The value of $\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ} = $

If $\sin A + \sin B = C$ and $\cos A + \cos B = D$,then the value of $\sin (A + B) = $

If $\alpha + \beta - \gamma = \pi,$ then $\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = $

$A$ ladder having a length of $16\,m$ is leaning against a wall at an angle of $60^{\circ}$ with the ground. Find the distance between the wall and the foot of the ladder (in $m$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module