Question

The value of $4\cos \frac{\pi }{10}-3\sec \frac{\pi }{10}-2\tan \frac{\pi }{10}$

• A. $1$
• B. $\sqrt{5}-1$
• C. $\sqrt{5}+1$
• D. $zero$

Answer 1

The correct option is A $1$

Given:-

$\sin x+{\sin }^{2}x=1$

$\Rightarrow \sin x=1-{\sin }^{2}x$

$\Rightarrow \sin x=\cos 2x$

Now

${\cos }^{8}x+2{\cos }^{6}x+{\cos }^{4}x$

$={\left({\cos }^{2}x\right)}^4+2{\left({\cos }^{2}x\right)}^3+{\left({\cos }^{2}x\right)}^2$

$={\sin }^{4}x+2{\sin }^{3}x+{\sin }^{2}x[\because {\cos }^{2}x=\sin x]$

$={\sin }^{4}x+{\sin }^{3}x+{\sin }^{3}x+{\sin }^{2}x$

$={\sin }^{2}x({\sin }^{2}x+\sin x)+\sin x({\sin }^{2}x+\sin x)$

$={\sin }^{2}x\left(1\right)+\sin x\left(1\right)[\because {\sin }^{2}x+\sin x=1]$

$={\sin }^{2}x+\sin x$

$=1$