Question

The value of $\frac{1-{\tan }^2{45}^{\circ }}{1+{\tan }^2{45}^{\circ }}=$

• A. $\tan {90}^{\circ }$
• B. $1$
• C. $\sin {45}^{\circ }$
• D. $0$

Answer 1

The correct option is D $0$

We know that $cos2\theta =\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }$

Then,

$\frac{1-ta{n}^2{45}^{\circ }}{1+ta{n}^2{45}^{\circ }}=cos(2\times {45}^{\circ })=cos{90}^{\circ }$

$=0$

$\therefore \frac{1-ta{n}^2{45}^{\circ }}{1+ta{n}^2{45}^{\circ }}=0$