Question

Match the entries of Column - I and Column - II.

	Column - I		Column - II
a	If 4 $si{n}^{-1}x+co{s}^{-1}x=\pi$, then x equals	1	ab
b	If $\angle C={90}^0$, then the value of $ta{n}^{-1}$ $\frac{a}{b+c}$ + $ta{n}^{-1}$ $\frac{b}{c+a}$ is	2	$\pi$
c	$ta{n}^{-1}$ 1 + $ta{n}^{-1}$ 2 + $ta{n}^{-1}$ 3 is	3	$\pi$/4
d	If $se{c}^{-1}$ $\frac{x}{a}$ - $se{c}^{-1}$ $\frac{x}{b}$ = $se{c}^{-1}$ b - $se{c}^{-1}$ a, then x equals	4	1/2

• A. a-4, b-3, c-2, d-1
• B. a-1, b-3, c-2, d-4
• C. a-4, b-3, c-1, d-2
• D. a-3, b-4, c-2, d-1

Answer 1

The correct option is A a-4, b-3, c-2, d-1

(a) $\to$ (4).
4 $si{n}^{-1}$ x + $(\frac{\pi }{2}-si{n}^{-1}x)$ = x
$\therefore$ 3$si{n}^{-1}$ x = $\frac{\pi }{2}$ or x = $sin\frac{\pi }{6}=\frac{1}{2}$
(b) $\to$ (3).
Use Rule 5 and $a^2$ + $b^2$ = $c^2$, $ta{n}^{-1}$ 1 = $\pi /4$
(c) $\to$ (2).
$ta{n}^{-1}$ 2 + $ta{n}^{-1}$ 3 = $\pi$ + $ta{n}^{-1}$ $\frac{2+3}{1-2.3}$
= $\pi$ - $ta{n}^{-1}$ 1
$\therefore$ L.H.S = $ta{n}^{-1}$ 1 +$ta{n}^{-1}$ 2 + $ta{n}^{-1}$ 3 = $\pi$
(d) $\to$ (1)