\(\frac{1-Tan^2\,45^\circ}{1+Tan^2\,45^\circ}\) can not be expressed

1.	\(\frac{1-Tan^2\,45^\circ}{1+Tan^2\,45^\circ}\) can not be expressed as ………A) Sin 0°B) cos 90° C) Sin 0° Cos 90° D) Sin 60°
Answer» Correct option is: D) Sin 60° \(\frac{1-Tan^2 45^\circ}{1+Tan^2 45^\circ}\) = \(\frac {1-1^2}{1+1^2} = \frac {1-1}{1+1} = \frac 02 = 0\) (\(\because\) tan \(45^\circ\) = 1) \(\because\) sin \(0^\circ\) = 0, cos \(90^\circ\) = 0 and sin \(0^\circ\). cos \(90^\circ\) = 0 \(\times\) 0 = 0 But sin \(60^\circ\) = \(\frac {\sqrt3}{2} \neq 0\) \(\therefore\) \(\frac{1-Tan^2 45^\circ}{1+Tan^2 45^\circ}\) can not be expressed as sin \(60^\circ\). Correct option is: D) Sin 60°

\(\frac{1-Tan^2\,45^\circ}{1+Tan^2\,45^\circ}\) can not be expressed as ………A) Sin 0°B) cos 90° C) Sin 0° Cos 90° D) Sin 60°

Answer»

Correct option is: D) Sin 60°

\(\frac{1-Tan^2 45^\circ}{1+Tan^2 45^\circ}\) = \(\frac {1-1^2}{1+1^2} = \frac {1-1}{1+1} = \frac 02 = 0\) (\(\because\) tan \(45^\circ\) = 1)

\(\because\) sin \(0^\circ\) = 0, cos \(90^\circ\) = 0

and sin \(0^\circ\). cos \(90^\circ\) = 0 \(\times\) 0 = 0

But sin \(60^\circ\) = \(\frac {\sqrt3}{2} \neq 0\)

\(\therefore\) \(\frac{1-Tan^2 45^\circ}{1+Tan^2 45^\circ}\) can not be expressed as sin \(60^\circ\).

Correct option is: D) Sin 60°