Practicing Success
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)$ |
Statement 1 is True, Statement 2 is true; Statement 2 is a correct explanation for Statement 1. Statement 1 is True, Statement 2 is True; Statement 2 is not a correct explanation for Statement 1. Statement 1 is True, Statement 2 is False. Statement 1 is False, Statement 2 is True. |
Statement 1 is True, Statement 2 is False. |
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. |