Question : If $\tan ^2 \alpha=3+Q^2$, then $\sec \alpha+\tan ^3 \alpha \operatorname{cosec} \alpha=?$
Option 1: $\left(3+Q^2\right)^{\frac{3}{2}}$
Option 2: $\left(7+Q^2\right)^{\frac{3}{2}}$
Option 3: $\left(5-Q^2\right)^{\frac{3}{2}}$
Option 4: $\left(4+Q^2\right)^{\frac{3}{2}}$
Correct Answer: $\left(4+Q^2\right)^{\frac{3}{2}}$
Solution :
Consider, $\sec \alpha+\tan ^3 \alpha \operatorname{cosec} \alpha$
= $\frac{1}{\cos\alpha}+\frac{\sin^3\alpha}{\cos^3\alpha}\times\frac{1}{\sin\alpha}$
= $\frac{1}{\cos\alpha}+\frac{\sin^2\alpha}{\cos^3\alpha}$
= $\frac{\cos^2\alpha+\sin^2\alpha}{\cos^3\alpha}$
= $\frac{1}{\cos^3\alpha}$ [Using $\cos^2\alpha+\sin^2\alpha = 1$] ....................(1)
Now consider, $\tan ^2 \alpha=3+Q^2$
⇒ $\tan\alpha = \sqrt{3+Q^2}$
Let base ($b$) = 1 and perpendicular ($p$) = $\sqrt{3+Q^2}$
Applying Pythagoras Theorem,
$h^2=p^2+b^2$ [where $h,p,$ and $b$ are hypotenuse, perpendicular and base respectively.]
⇒ $h=\sqrt{1+3+Q^2}$
⇒ $h=\sqrt{4+Q^2}$
⇒ $\cos\alpha = \frac{b}{h}=\frac{1}{\sqrt{4+Q^2}}$
Putting in (1), we get,
$\frac{1}{\cos^3\alpha}=\frac{1}{(\frac{1}{\sqrt{4+Q^2}})^3} = (4+Q^2)^{\frac{3}{2}}$
Hence, the correct answer is $(4+Q^2)^{\frac{3}{2}}$.
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