Question

If $\pi ={180}^o$ and $A=\frac{\pi }{6}$, prove that $\frac{(1-\cos A)(1+\cos A)}{(1-\sin A)(1+\sin A)}=\frac{1}{3}$

Answer 1

If $\pi ={180}^0$ then $A=\frac{\pi }{6}=\frac{{180}^0}{6}={30}^0$.

Let usfirstfind the value of left hand side (LHS) that is $\frac{(1-\cos A\left)\right(1+\cos A)}{(1-\sin A)(1+\sin A)}$ with $A={30}^0$ as shown below:

$\frac{(1-\cos A\left)\right(1+\cos A)}{(1-\sin A)(1+\sin A)}=\frac{1-{\cos }^{2}A}{1-{\sin }^{2}A}(\because (a+b\left)\right(a-b)=a^2-b^2)=\frac{{\sin }^{2}A}{{\cos }^{2}A}(\because {\sin }^{2}x+{\cos }^{2}x=1)={\tan }^{2}A(\because \frac{\sin A}{\cos A}=\tan A)=(\tan {30}^0{)}^2(\because A={30}^0)={\left(\frac{1}{\sqrt{3}}\right)}^2(\because \tan {30}^0=\frac{1}{\sqrt{3}})=\frac{1}{3}=RHS$

Since LHS=RHS,

Hence, $\frac{(1-\cos A\left)\right(1+\cos A)}{(1-\sin A)(1+\sin A)}=\frac{1}{3}$.