Question

The expression $({\tan }^{4}x+2{\tan }^{2}x+1){\cos }^{2}x$ at $x=\frac{\pi }{12}$ can be equal to

• A. $4(2-\sqrt{3})$
• B. $4(\sqrt{2}+1)$
• C. $16\frac{{\cos }^2\pi }{12}$
• D. $16\frac{{\sin }^2\pi }{12}$

Answer 1

Answer: (A), (D)

The correct options are
A $4(2-\sqrt{3})$
D $16\frac{{\sin }^2\pi }{12}$

$a=({\tan }^{4}x+2{\tan }^{2}x+1){\cos }^{2}x\Rightarrow a={(1+{\tan }^{2}x)}^2\frac{4{\sin }^{2}x}{4{\tan }^{2}x}\Rightarrow a=4{\left(\frac{1+{\tan }^{2}x}{2\tan x}\right)}^2\left(\frac{1-\cos 2x}{2}\right)\Rightarrow a=2\frac{(1-\cos 2x)}{{\sin }^{2}2x}$
Substitute $x=\frac{\pi }{12}$
$a=2\frac{(1-\cos \frac{\pi }{6})}{{\sin }^2\frac{\pi }{6}}=\frac{4{\sin }^2\frac{\pi }{12}}{\frac{1}{4}}=16{\sin }^2\frac{\pi }{12}a=2\frac{(1-\cos \frac{\pi }{6})}{{\sin }^2\frac{\pi }{6}}=\frac{2(1-\frac{\sqrt{3}}{2})}{\frac{1}{4}}=4(2-\sqrt{3})$
Ans: A,D