Statement-1: If $a^2 + b^2 =c^2 , c ≠ 0 , ab ≠ 0,$ then the non-zero solution of the equation $sin^{-1}\frac{ax}{c}+sin^{-1}\frac{bx}{c}=sin^{-1}x$ is , ±1.
Statement-2: $ sin^{-1}x +sin^{-1}y =sin^{-1}(x + y)$
Answer & explanation
Correct answer: option 3
Clearly, statement-2 is not true.
We have,
$c^2 = a^2 +b^2. $ So, let a = c cos $\alpha $ and b = c sin $\alpha $. Then,
$sin^{-1}\frac{ax}{c}+sin^{-1}\frac{bx}{c}= sin^{-1}x $
$⇒ sin^{-1} (x cos \alpha ) + sin^{-1} ( x sin \alpha ) = sin^{-1} x $
$⇒ sin^{-1} \begin{Bmatrix} x cos \alpha \sqrt{1-x^2 sin^ \alpha} + x sin \alpha \sqrt{1-x^2 cos^2 \alpha }\end{Bmatrix}= sin^{-1}x$
$⇒ cos \alpha \sqrt{1-x^2 sin^2 \alpha} + sin \alpha \sqrt{1-x^2 cos^2 \alpha } = 1 $ [∵ x ≠ 0]
Clearly, x = ± 1 satisfies this equation.