Question

Match the following
List - I List-II
1. $\sin {50}^0-\sin {70}^0+\sin {10}^0$ $a.4$
2. $\sqrt{3}cosec{20}^0-\sec {20}^0$ $b.-2\sqrt{3}$
3. $\tan {15}^0-\cot {15}^0$ $c.0$
4 $\sin {15}^0\cos {15}^0$ $d.\frac{1}{4}$

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

Answer 1

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

1. $\sin {50}^o-\sin {70}^o+\sin {10}^o=-2\cos {60}^o\sin {10}^o+\sin {10}^o$

$=\sin {10}^o-\sin {10}^o[\because \sin A-\sin B=2\cos (\frac{A+B}{2}\left)\sin \right(\frac{A-B}{2}\left)\right]$

$=0$

2. $\sqrt{3}cosec{20}^o-\sec {20}^o=\frac{\sqrt{3}\cos {20}^o-\sin {20}^o}{\cos {20}^o\sin {20}^o}$

$=2\frac{\sin (60-20)}{\cos {20}^o\sin 20}[\because \sin A\cos B-\cos A\sin B=\sin (A-B\left)\right]$

$=4\frac{\sin 40}{\sin 40}=4$

3. $\tan 15-cot15=\frac{{\sin }^2{15}^o-{\cos }^2{15}^o}{\cos {15}^o\sin {15}^o}$

$=\frac{-2\cos {30}^o}{\sin {30}^o}=-2\sqrt{3}[\because {\cos }^2\theta -{\sin }^2\theta =\cos 2\theta ]$

4. $\sin {15}^o\cos {15}^o=\frac{\sin {30}^o}{2}=\frac{1}{4}[\because \sin 2\theta =2\sin \theta \cos \theta ]$