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\(\frac{1-tan^2\,30^\circ}{1+tan^2\,30^\circ}\) = ……..A) 1/2B) 1/√2C) √3/2D) 1 |
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Answer» Correct option is: A) \(\frac{1}{2}\) \(\frac{1-tan^2 30^\circ}{1+tan^2 30^\circ} = \frac {1-(\frac 1{\sqrt3})^2}{1+ (\frac 1{\sqrt3})^2}\) ( \(\because\) tan \(30^\circ\) = \(\frac 1{\sqrt3}\)) = \(\frac {1-\frac 13}{1+ \frac 13} = \frac {\frac {3-1}3}{\frac {3+1}3} = \frac 24 = \frac 12\) Correct option is: A) \(\frac{1}{2}\) |
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