Question

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

	Column - I		Column - II
a	x = $cose{c}^2$ $\left(co{t}^{-1}3\right)$ - $se{c}^2$ $\left(ta{n}^{-1}2\right)$	p	x = 2
b	$ta{n}^{-1}$ x + $ta{n}^{-1}$ $\frac{1}{y}$ = $\left(ta{n}^{-1}3\right)$ and $y^2$ + y - 56 = 0	q	x = 5
c	$co{s}^{-1}$ x = $ta{n}^{-1}$ y and $y^2$ = 3	r	$x=\frac{1}{2}$
d	$si{n}^{-1}$ $\left(tan\frac{\pi }{4}\right)$ - $si{n}^{-1}$ $\sqrt{\frac{3}{y}}$ = $\frac{\pi }{6}$ and $x^2$ = y	s	x = - $\frac{1}{2}$

• A. $a-\left(q\right),b-(p,q),c-(r,s),d-\left(p\right)$
• B. $a-\left(q\right),b-(p,),c-(r,s),d-\left(p\right)$
• C. $a-\left(q\right),b-(p,q),c-(r,),d-\left(p\right)$
• D. $a-\left(p\right),b-(p,q),c-(r,s),d-\left(p\right)$

Answer 1

The correct option is A $a-\left(q\right),b-(p,q),c-(r,s),d-\left(p\right)$

(a)$\to$(q)
R.H.S = $[1+co{t}^2(co{t}^{-1}3\left)\right]-[1+ta{n}^2(ta{n}^{-1}2\left)\right]=3^2-2^2=5$
x = 5
(b)$\to$(p,q)
$\frac{x+\left(1/y\right)}{1-\left(x/y\right)}$ = 3 provided (x/y) < 1, i.e., x<y
xy + 1 = 3(y-x)
Also $y^2$ + y - 56 = 0 $\Rightarrow$ (y + 8)(y - 7) = 0
y = -8,7
When y = -8, x = -25/-5 = 5
x/y = (-5/8)<1
When y = 7, x = 20/10 = 2
x/y = 2/7 <1
(c)$\to$(r,s)
$ta{n}^{-1}=\Theta$
y = $tan\Theta$
1 + $ta{n}^2\Theta =1+y^2$
$se{c}^2\Theta =1+y^2$
$cos\Theta =\frac{1}{\sqrt{1+y^2}}$
$\Theta =co{s}^{-1}x$
x = $\frac{1}{\sqrt{1+y^2}}$ = $\frac{1}{\sqrt{1+3}}$ = 1/$\sqrt{4}$ = $\pm 1/2$
(d)$\to$(p)
$si{n}^{-1}-\left(\pi /6\right)=si{n}^{-1}\sqrt{3/y}$ and $x^2$ = y
$\left(\pi /2\right)-\left(\pi /6\right)=si{n}^{-1}\sqrt{3/y}$ or $\sqrt{3/y}=sin{60}^0=\sqrt{3}/2$
$\sqrt{y}$ = 2 or y = 4 = $x^2$ $\Rightarrow$ x = $\pm$2