Question

The value of the expression $({\tan }^{4}x+2{\tan }^{2}x+1){\cos }^{2}x$, when $x=\frac{\pi }{12}$ is equal to

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

Answer 1

Answer: (A), (D)

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

$({\tan }^{4}x+2{\tan }^{2}x+1){\cos }^{2}x=\frac{(1+{\tan }^{2}x{)}^2}{{\sec }^{2}x}=\frac{(1+{\tan }^{2}x{)}^2}{1+{\tan }^{2}x}=1+{\tan }^{2}x$
Putting $x=\frac{\pi }{12}$

$=1+{\tan }^2\frac{\pi }{12}=1+{(2-\sqrt{3})}^2=4(2-\sqrt{3})$

Also,
$16{\cos }^2\frac{\pi }{12}=4{\left(\frac{\sqrt{3}+1}{\sqrt{2}}\right)}^2=4(2+\sqrt{3})16{\sin }^2\frac{\pi }{12}=4{\left(\frac{\sqrt{3}-1}{\sqrt{2}}\right)}^2=4(2-\sqrt{3})$