Statement-1: If a is twice the tangent of the arithmetic mean of $sin^{-1}x$ and $cos^{-1}x$, b is the geometric mean of tan x and cot x, then $x^2 - ax + b = 0 ⇒ x = 1 $
Statement-2 : $ tan \left(\frac{sin^{-1}x + cos^{-1}x}{2}\right)=1$
Answer & explanation
Correct answer: option 1
It is given that
$a = 2 tan \left(\frac{sin^{-1}x + cos^{-1}x}{2}\right)$ and $ b = \sqrt{tan x × cot x }$
$⇒ a = 2 tan \frac{\pi}{4} = 2 $ and $ b = 1$
$∴ x^2 - ax + b = 0 ⇒x^2 - 2x + 1 = 0 ⇒ (x-1)^2 = 0 ⇒ x= 1 $
So, statement -1 is true
$tan \left(\frac{sin^{-1}x + cos^{-1}x}{2}\right)= tan \frac{\pi}{4} = 1 $
So, statement-2 is also true.