Question

A quadratic equation whose roots are $\tan 22\frac{1^0}{2}$ and $\cot 22\frac{1^0}{2}$ is:

• A. $x^2-2\sqrt{2}x+1=0$
• B. $2x^2-\sqrt{2}x+1=0$
• C. $x^2+2\sqrt{2}x-1=0$
• D. $x^2-2\sqrt{2}x-1=0$

Answer 1

Answer: (A)

$x^2-2\sqrt{2}x+1=0$

Sum of the roots $=\tan 22{\frac{1}{2}}^0+\cot 22{\frac{1}{2}}^0$

$=\frac{\sin 22{\frac{1}{2}}^0}{\cos 22{\frac{1}{2}}^0}+\frac{\cos 22{\frac{1}{2}}^0}{\sin 22{\frac{1}{2}}^0}$

$=\frac{{\sin }^2\left(22{\frac{1}{2}}^0\right)+{\cos }^2\left(22{\frac{1}{2}}^0\right)}{\sin 22{\frac{1}{2}}^0\cos 22{\frac{1}{2}}^0}$

$=\frac{1}{\sin 22{\frac{1}{2}}^0\cos 22{\frac{1}{2}}^0}$

$=\frac{2}{2\sin 22{\frac{1}{2}}^0\cos 22{\frac{1}{2}}^0}$

$=\frac{2}{\sin {45}^0}$

$=2\sqrt{2}$

Product of roots $=\tan 22{\frac{1}{2}}^0\times \cot 22{\frac{1}{2}}^0=1$

So, the quadratic equation is $x^2-\left(sumofroots\right)x+\left(productofroots\right)=0$

$⟹x^2-2\sqrt{2}x+1=0$