Question

If $p=\sec \alpha -\tan \alpha ,q=\text{cosec}\alpha +\cot \alpha$. Find $q$ in terms of $p$.

• A. ${}^{\frac{p+1}{p-1}}$
• B. $\frac{1+p}{1-p}$
• C. $\frac{p-1}{p+1}$
• D. $\frac{1-p}{1+p}$

Answer 1

The correct option is B $\frac{1+p}{1-p}$

$\frac{1+p}{1-p}=\frac{1+\cos \alpha -\sin \alpha }{-1+\cos \alpha +\sin \alpha }$

$=\frac{2{\cos }^2\frac{\alpha }{2}-2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}{-2{\sin }^2\frac{\alpha }{2}+2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}$

$\frac{\cos \frac{\alpha }{2}(\cos \frac{\alpha }{2}-\sin \frac{\alpha }{2})}{\sin \frac{\alpha }{2}(-\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2})}=\frac{{\cos }^2\frac{\alpha }{2}}{\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}$

Multiply numerator and denominator by $2$, we get

$=\frac{{2\cos }^2\frac{\alpha }{2}}{2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}=\frac{1+\cos \alpha }{\sin \alpha }=\text{cosec}\alpha +\cot \alpha =q$

Hence, $q=\frac{1+p}{1-p}$