\(cos(2 tan^{-1}\frac{1}{2})=?\)A. \(\frac{3}{5}\)B. \(\frac{4}{5}\)

1.	\(cos(2 tan^{-1}\frac{1}{2})=?\)A. \(\frac{3}{5}\)B. \(\frac{4}{5}\)C. \(\frac{7}{8}\)D. none of these
Answer» Correct Answer is (A) \(\frac{3}{5}\) Let , x = \(cos(2 tan^{-1}\frac{1}{2})\) ⇒ x = cos (tan^-1\(\frac{1}{3}\) + 2 tan^-1\(\frac{1}{3}\)) Since we know that tan^-1x + tan^-1 y = \(tan^{-1}\frac{x+y}{1-xy}\) ⇒ tan^-1 1 + tan^-1 \(\frac{1}{3}\)= tan^-1\((\frac{\frac{1}{2}+\frac{1}{2}}{1-(\frac{1}{2}\times\frac{1}{2})})\)= \(tan^{-1}1\)\(\frac{4}{3}\) Now, let y = \(tan^{-1}1\)\(\frac{4}{3}\) ⇒ tan y = \(\frac{4}{3}\) = \(\frac{\text{opposite side}}{\text{adjacent side}}\) By pythagorus theroem , (Hypotenuse )²= (opposite side )² + (adjacent side )² Therefore, Hypotenuse = 5 ⇒ \(cos(tan^{-1}\frac{4}{3})\) = cos y = \(\frac{3}{5}\)

\(cos(2 tan^{-1}\frac{1}{2})=?\)A. \(\frac{3}{5}\)B. \(\frac{4}{5}\)C. \(\frac{7}{8}\)D. none of these

Answer»

Correct Answer is (A) \(\frac{3}{5}\)

Let , x = \(cos(2 tan^{-1}\frac{1}{2})\)

⇒ x = cos (tan^-1\(\frac{1}{3}\) + 2 tan^-1\(\frac{1}{3}\))

Since we know that tan^-1x + tan^-1 y = \(tan^{-1}\frac{x+y}{1-xy}\)

⇒ tan^-1 1 + tan^-1 \(\frac{1}{3}\)= tan^-1\((\frac{\frac{1}{2}+\frac{1}{2}}{1-(\frac{1}{2}\times\frac{1}{2})})\)= \(tan^{-1}1\)\(\frac{4}{3}\)

Now, let y = \(tan^{-1}1\)\(\frac{4}{3}\)

⇒ tan y = \(\frac{4}{3}\) = \(\frac{\text{opposite side}}{\text{adjacent side}}\)

By pythagorus theroem ,

(Hypotenuse )²= (opposite side )² + (adjacent side )²

Therefore, Hypotenuse = 5

⇒ \(cos(tan^{-1}\frac{4}{3})\) = cos y = \(\frac{3}{5}\)