Question

Solution of the quadratic equation $x^2-\left(\tan \frac{a}{2}+\cot \frac{a}{2}\right)x+\tan \frac{a}{2}\times \cot \frac{a}{2}=0$ is:

• A. $\frac{1+\cos a}{\sin a}$
• B. $\frac{1-\cos a}{\sin a}$
• C. $\frac{\cos a}{1+\sin a}$
• D. $\frac{\cos a}{1-\sin a}$

Answer 1

Answer: (A)

$\frac{1+\cos a}{\sin a}$

Given quadratic equation $x^2-(\tan \frac{a}{2}+\cot \frac{a}{2})x+\tan \frac{a}{2}\times \cot \frac{a}{2}=0$

$\Rightarrow x^2-(\frac{\sin \frac{a}{2}}{\cos \frac{a}{2}}+\frac{\cos \frac{a}{2}}{\sin \frac{a}{2}})x+1=0$

$\Rightarrow x^2-\left(\frac{{\sin }^2\frac{a}{2}+{\cos }^2\frac{a}{2}}{\cos \frac{a}{2}\times \sin \frac{a}{2}}\right)x+1=0$

$\Rightarrow x^2-\left(\frac{1}{\cos \frac{a}{2}\times \sin \frac{a}{2}}\right)x+1=0$

$\Rightarrow x^2-(2\times \frac{1}{\sin a})x+1=0$

$\Rightarrow x^2-2\times x\times \frac{1}{\sin a}+\frac{1}{{\sin }^{2}a}-\frac{1}{{\sin }^{2}a}+1=0$

$\Rightarrow {(x-\frac{1}{\sin a})}^2=\frac{1}{{\sin }^{2}a}-1$

$\Rightarrow {(x-\frac{1}{\sin a})}^2=\frac{1-{\sin }^{2}a}{{\sin }^{2}a}$

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

$\Rightarrow x-\frac{1}{\sin a}=\sqrt[2]{2\frac{{\cos }^{2}a}{{\sin }^{2}a}}$

$\Rightarrow x-\frac{1}{\sin a}=\frac{\cos a}{\sin a}$

$\Rightarrow x=\frac{1+\cos a}{\sin a}$