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Question:

If $\tan \theta=\frac{1}{\sqrt{7}}$, then $\frac{\left(\operatorname{cosec}^{2} \theta-\sec ^{2} \theta\right)}{\left(\operatorname{cosec}^{2} \theta+\sec ^{2} \theta\right)}=?=?$

(a) $\frac{-2}{3}$

(b) $\frac{-3}{4}$

(c) $\frac{2}{3}$

(d) $\frac{3}{4}$

 

Solution:

(d) $\frac{3}{4}$

$=\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}$

$=\frac{\sin ^{2} \theta\left(\frac{1}{\sin ^{2} \theta}-\frac{1}{\cos ^{2} \theta}\right)}{\sin ^{2} \theta\left(\frac{1}{\sin ^{2} \theta}+\frac{1}{\cos ^{2} \theta}\right)} \quad\left[\right.$ Multiplying the numerator and denominator by $\left.\sin ^{2} \theta\right]$

$=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$

$=\frac{1-\frac{1}{7}}{1+\frac{1}{7}}=\frac{6}{8}=\frac{3}{4}$

 

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