Statement-1: If $-1 ≤ x ≤ 1,$ then
$sin^{-1} (-x)= -sin^{-1} x $ and $ cos^{-1} (-x) = \pi - cos^{-1} x $
Statement-2: If $-1 ≤ x ≤ 1,$ then
$cos^{-1} x = 2 sin^{-1} \sqrt{\frac{1-x}{2}}= 2cos^{-1} \sqrt{\frac{1+x}{2}}$
Answer & explanation
Correct answer: option 2
Clearly, statement-1 is true.
Putting x = cos $\theta $, we get
$2 sin^{-1} \sqrt{\frac{1-x}{2}}= 2cos^{-1} \sqrt{\frac{1+x}{2}}= \theta = cos^{-1}x$
So, statement-2 is also true.