Question

If $tanx+ta{n}^{2}x+ta{n}^{3}x=1$ then the value of $2co{s}^{6}x-2co{s}^{4}x+co{s}^{2}x$ equal to

• A. $\frac{1}{2}$
• B. $2$
• C. $1$
• D. none of these

Answer 1

The correct option is A $\frac{1}{2}$

$\tan x+{\tan }^{2}x+{\tan }^{3}x=1\Rightarrow {\cos }^{2}x\sin x+{\sin }^{3}x+\cos x{\sin }^{2}x={\cos }^{3}x$

$\Rightarrow \sin x({\cos }^{2}x+{\sin }^{2}x)={\cos }^{3}x-\cos x{\sin }^{2}x$

$\Rightarrow \sin x={\cos }^{3}x-\cos x(1-{\cos }^{2}x)=2{\cos }^{3}x-\cos x$

On Squaring,

${\sin }^{2}x=4{\cos }^{6}x+{\cos }^{2}x-4{\cos }^{4}x$

$\Rightarrow (1-{\cos }^{2}x)=4{\cos }^{6}x+{\cos }^{2}x-4{\cos }^{4}x$

$\Rightarrow 2{\cos }^{6}x-2{\cos }^{4}x+{\cos }^{2}x=\frac{1}{2}$

So, Option $A$ is correct answer.