Question

The value of the expression $\frac{ta{n}^2{20}^0-si{n}^2{20}^0}{ta{n}^2{20}^0.si{n}^2{20}^0}$ is

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

Answer 1

The correct option is A $1$

Given expression is $\frac{{\tan }^2{20}^0-{\sin }^2{20}^0}{{\tan }^2{20}^0.{\sin }^2{20}^0}$

We know that, $\tan \theta =\frac{\sin \theta }{\cos \theta }$

Hence, ${\tan }^2\theta =\frac{{\sin }^2\theta }{{\cos }^2\theta }$

$\Rightarrow \frac{{\tan }^2{20}^0-{\sin }^2{20}^0}{{\tan }^2{20}^0.{\sin }^2{20}^0}=\frac{\frac{{\sin }^2{20}^0}{{\cos }^2{20}^0}-{\sin }^2{20}^0}{\frac{{\sin }^2{20}^0}{{\cos }^2{20}^0}{\sin }^2{20}^0}$
$=\frac{{\sin }^2{20}^0-{\sin }^2{20}^0{\cos }^2{20}^0}{{\sin }^4{20}^0}$

$=\frac{{\sin }^2{20}^0(1-{\cos }^2{20}^0)}{{\sin }^4{20}^0}$
We also know that, ${\sin }^2\theta +{\cos }^2\theta =1$

Hence, $1-{\cos }^2\theta ={\sin }^2\theta$
So, $\frac{{\sin }^2{20}^0{\sin }^2{20}^0}{{\sin }^4{20}^0}$

$=\frac{{\sin }^4{20}^0}{{\sin }^4{20}^0}$

$=1$

Hence, $\frac{{\tan }^2{20}^0-{\sin }^2{20}^0}{{\tan }^2{20}^0.{\sin }^2{20}^0}=1$.