Question

If cos(2tan⁻¹x)=1/2, then the value of x is.......,Select Proper option from the given options.
(a) 1/√3
(b) 1-√3
(c) 1- 1/√3
(d) √3

Answer 1

Dear Student,

Answer:Option a is correct. x= 1/√3

Solution:

1) First convert 2tan⁻¹x into cos⁻¹x, so that cos cancels cos⁻¹x
$2 {tan}^{ - 1} x = {cos}^{ - 1} ( \frac{1 - {x}^{2} }{1 + {x}^{2} } ) \\ \\ \\ cos (\: 2 {tan}^{ - 1} x) = \\ \\ cos( {cos}^{ - 1} ( \frac{1 - {x}^{2} }{1 + {x}^{2} } ) = \frac{1}{2} \\ \\ ( \frac{1 - {x}^{2} }{1 + {x}^{2} } ) = \frac{1}{2} \\ \\ 2 - 2 {x}^{2} = 1 + {x}^{2} \\ \\ - 2 {x}^{2} - {x}^{2} = 1 - 2 \\ \\ - 3 {x}^{2} = - 1 \\ 3 {x}^{2} = 1 \\ {x}^{2} = \frac{1}{3} \\ \\ x = + - \sqrt{ \frac{1}{3} } \\ \\ x = - \frac{1}{ \sqrt{3} } \\ \\ x = - \frac{1}{ \sqrt{3} }$
So, one of the value is 1/√3

Option a is correct

Hope it helps you.

Answer 2

we have to find the value of x where $cos(2tan^{-1}x)=1/2$.

Let $2tan^{-1}x=A$
tanA/2 = x
we know, $cos2\theta=\frac{1-tan^2\theta}{1+tan^2\theta}$

so, cosA = (1 - tan²A/2)/(1 + tan²A/2)

= (1 - x²)/(1 + x²)

so, $\frac{1-x^2}{1+x^2}=cosA$......(1)

now, $cos(2tan^{-1}x)=1/2$

=> cosA = 1/2

from equation (1),

=> (1 - x²)/(1 + x²) = 1/2

=> 2(1 - x²) = (1 + x²)

=> 2 - 1 = 3x²

=> x = ±1/√3

hence option (a) is correct.