MCQ
If $(\cos\theta+\sec\theta)=\frac{5}{2}$ then $(\cos^2\theta+\sec^2\theta)=?$
- A$\frac{21}{4}$
- ✓$\frac{17}{4}$
- C$\frac{29}{4}$
- D$\frac{33}{4}$
✓
Answer
Correct option: B.
$\frac{17}{4}$
$(\cos\theta+\sec\theta)=\frac{5}{2}$
$\Rightarrow(\cos\theta+\sec\theta)^2=\Big(\frac{5}{2}\Big)^2$
$\Rightarrow\cos^2\theta+\sec^2\theta+2\cos\theta\sec\theta=\frac{25}{4}$
$\Rightarrow\cos^2\theta+\sec^2\theta+2\cos\theta\times\frac{1}{\cos\theta}=\frac{25}{4}$
$\Rightarrow\cos^2\theta+\sec^2\theta+2=\frac{25}{4}$
$\Rightarrow\cos^2\theta+\sec^2\theta=\frac{25}{4}-2$
$\Rightarrow\cos^2\theta+\sec^2\theta=\frac{25-8}{4}$
$\Rightarrow\cos^2\theta+\sec^2\theta=\frac{17}{4}$
$\Rightarrow(\cos\theta+\sec\theta)^2=\Big(\frac{5}{2}\Big)^2$
$\Rightarrow\cos^2\theta+\sec^2\theta+2\cos\theta\sec\theta=\frac{25}{4}$
$\Rightarrow\cos^2\theta+\sec^2\theta+2\cos\theta\times\frac{1}{\cos\theta}=\frac{25}{4}$
$\Rightarrow\cos^2\theta+\sec^2\theta+2=\frac{25}{4}$
$\Rightarrow\cos^2\theta+\sec^2\theta=\frac{25}{4}-2$
$\Rightarrow\cos^2\theta+\sec^2\theta=\frac{25-8}{4}$
$\Rightarrow\cos^2\theta+\sec^2\theta=\frac{17}{4}$
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