Question

$$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{a.}\sin ({410}^{\circ }-A)\cos ({400}^{\circ }+A)+\cos ({410}^{\circ }-A)\sin ({400}^{\circ }+A)& \text{p. -1}\\ \text{b.}\frac{{\cos }^2{1}^{\circ }-{\cos }^2{2}^{\circ }}{2\sin 3^{\circ }\sin 1^{\circ }}\text{is equal to}& \text{q. 1}\\ \text{c.}\sin (-{870}^{\circ })+\text{cosec}(-{660}^{\circ })+\tan (-{855}^{\circ })+2\cot \left({840}^{\circ }\right)+\cos \left({480}^{\circ }\right)+\sec \left({900}^{\circ }\right)& \text{r.}\frac{1}{2}\end{array}$$

Which of the following is the CORRECT combination ?

• A. $a\to p,b\to q,c\to r$
• B. $a\to r,b\to q,c\to p$
• C. $a\to q,b\to r,c\to p$
• D. $a\to r,b\to p,c\to q$

Answer 1

The correct option is C $a\to q,b\to r,c\to p$

$a.$ $\sin ({410}^{\circ }-A)\cos ({400}^{\circ }+A)+\cos ({410}^{\circ }-A)\sin ({400}^{\circ }+A)$
$=\sin ({410}^{\circ }-A+{400}^{\circ }+A)$
$=\sin \left({810}^{\circ }\right)$
$=\sin (4\pi +\frac{\pi }{2})$
$=\sin \frac{\pi }{2}$
$=1$


$b.$ $\frac{{\cos }^2{1}^{\circ }-{\cos }^2{2}^{\circ }}{2\sin 3^{\circ }\sin 1^{\circ }}$

$=\frac{(1-{\sin }^2{1}^{\circ })-(1-{\sin }^2{2}^{\circ })}{2\sin 3^{\circ }\sin 1^{\circ }}$

$=\frac{{\sin }^2{2}^{\circ }-{\sin }^2{1}^{\circ }}{2\sin 3^{\circ }\sin 1^{\circ }}$

$=\frac{\sin 3^{\circ }\cdot \sin 1^{\circ }}{2\sin 3^{\circ }\sin 1^{\circ }}$

$=\frac{1}{2}$


$c.$ $\sin (-{870}^{\circ })+\text{cosec}(-{660}^{\circ })+\tan (-{855}^{\circ })+2\cot \left({840}^{\circ }\right)+\cos \left({480}^{\circ }\right)+\sec \left({900}^{\circ }\right)$

$=-\sin ({810}^{\circ }+{60}^{\circ })-\text{cosec}({630}^{\circ }+{30}^{\circ })-\tan ({810}^{\circ }+{45}^{\circ })+2\cot ({810}^{\circ }+{30}^{\circ })+\cos ({450}^{\circ }+{30}^{\circ })+\sec ({810}^{\circ }+{90}^{\circ })$

$=-\cos {60}^{\circ }+\sec {30}^{\circ }+\cot {45}^{\circ }-2\tan {30}^{\circ }-\sin {30}^{\circ }-\text{cosec}{90}^{\circ }$

$=-\frac{1}{2}+\frac{2}{\sqrt{3}}+1-\frac{2}{\sqrt{3}}-\frac{1}{2}-1$
$=-1$