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