If $(\cos \theta+\sec \theta)=\frac{5}{2}$ then $\left(\cos ^{2} \theta+\sec ^{2} \theta\right)=?$
(a) $\frac{17}{4}$
(b) $\frac{21}{4}$
(c) $\frac{29}{4}$
(d) $\frac{33}{4}$
Given : $\cos \theta+\sec \theta=\frac{5}{2}$
$\cos \theta+\sec \theta=\frac{5}{2}$
Squaring both sides, we get
$\Rightarrow(\cos \theta+\sec \theta)^{2}=\left(\frac{5}{2}\right)^{2}$
$\Rightarrow \cos ^{2} \theta+\sec ^{2} \theta+2(\cos \theta)(\sec \theta)=\frac{25}{4}$
$\Rightarrow \cos ^{2} \theta+\sec ^{2} \theta+2(\cos \theta)\left(\frac{1}{\cos \theta}\right)=\frac{25}{4} \quad\left(\because \sec \theta=\frac{1}{\cos \theta}\right)$
$\Rightarrow \cos ^{2} \theta+\sec ^{2} \theta+2=\frac{25}{4}$
$\Rightarrow \cos ^{2} \theta+\sec ^{2} \theta=\frac{25}{4}-2$
$\Rightarrow \cos ^{2} \theta+\sec ^{2} \theta=\frac{25-8}{4}$
$\Rightarrow \cos ^{2} \theta+\sec ^{2} \theta=\frac{17}{4}$
Hence, the correct option is (a).
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