Question

If A = ${30}^{\circ }$, then $\cos 2A=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}={\cos }^{2}A-{\sin }^{2}A$
State true or false.

• A. True
• B. False

Answer 1

The correct option is A True

$\cos 2A={\cos }^{2}A-{\sin }^{2}A=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$
Given, $A={30}^{\circ }$
L.H.S: $\cos 2A=\cos 2\times {30}^0=\cos {60}^0=\frac{1}{2}$
Middle side: ${\cos }^{2}A-{\sin }^{2}A={\cos }^2{30}^0-{\sin }^2{30}^0=(\frac{\sqrt{3}}{2}{)}^2-(\frac{1}{2}{)}^2$

= $\frac{3}{4}-\frac{1}{4}$

= $\frac{1}{2}$
R.H.S: $\frac{1-{\tan }^{2}A}{1+ta{n}^{2}A}$

= $\frac{1-(\frac{1}{\sqrt{3}}{)}^2}{1+(\frac{1}{\sqrt{3}}{)}^2}$

= $\frac{3-1}{3+1}$

= $\frac{1}{2}$

Hence Proved