Question

Tha value of $\frac{{\sin }^{2}A}{1-\cos 2A}$ is

• A. $\frac{1}{2}$
• B. 0
• C. 1
• D. $\sqrt{3}$

Answer 1

Answer: (A)

$\frac{1}{2}$

The value of $\frac{si{n}^{2}A}{1-cos2A}$
We know that $cos2A=co{s}^{2}A-si{n}^{2}A$
Thus, $\frac{si{n}^{2}A}{1-cos2A}=\frac{si{n}^{2}A}{1-co{s}^{2}A+si{n}^{2}A}$
Using the identity $si{n}^{2}A+co{s}^{2}A=1$, we find that $1-co{s}^{2}A=si{n}^{2}A$
Thus, $\frac{si{n}^{2}A}{1-co{s}^{2}A+si{n}^{2}A}=\frac{si{n}^{2}A}{si{n}^{2}A+si{n}^{2}A}$
= $\frac{si{n}^{2}A}{2si{n}^{2}A}$= $\frac{1}{2}$