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

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$\frac{5}{6}$

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

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