Question

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(1)}& \frac{{\cos }^2{1}^{\circ }-{\cos }^2{2}^{\circ }}{2\sin 3^{\circ }\sin 1^{\circ }}\text{is equal to}& \left(p\right)& -1\\ \text{(2)}& \sin (-{870}^{\circ })+\text{cosec}(-{660}^{\circ })+\tan (-{855}^{\circ })& & \\ & +2\cot \left({840}^{\circ }\right)+\cos \left({480}^{\circ }\right)+\sec \left({900}^{\circ }\right)& \left(q\right)& \frac{1}{2}\\ \text{(3)}& \text{If}\cos \theta =\frac{4}{5}\text{where}\theta \in (\frac{3\pi }{2},2\pi )\text{and}& & \\ & \cos \alpha =\frac{3}{5}\text{where}\alpha \in (0,\frac{\pi }{2})\text{then}& & \\ & \cos (\theta -\alpha )\text{has the value equal to}& \left(r\right)& 0\end{array}$

Which of the following is the correct combination?

• A. $1\to p,2\to r,3\to q$
• B. $1\to p,2\to q,3\to r$
• C. $1\to q,2\to p,3\to r$
• D. $1\to r,2\to q,3\to p$

Answer 1

The correct option is C $1\to q,2\to p,3\to r$

$\left(1\right)$
$\frac{{\cos }^2{1}^{\circ }-{\cos }^2{2}^{\circ }}{2\sin 3^{\circ }\sin 1^{\circ }}$
$\Rightarrow \frac{\sin 3^{\circ }\sin 1^{\circ }}{2\sin 3^{\circ }\sin 1^{\circ }}=\frac{1}{2}$

$\left(2\right)$
$=\sin (-{870}^{\circ })+\text{cosec}(-{660}^{\circ })+\tan (-{855}^{\circ })+2\cot \left({840}^{\circ }\right)+\cos \left({480}^{\circ }\right)+\sec \left({900}^{\circ }\right)$
$=-\sin \left({870}^{\circ }\right)-\text{cosec}\left({660}^{\circ }\right)-\tan \left({855}^{\circ }\right)+2\cot ({810}^{\circ }+{30}^{\circ })+\cos ({450}^{\circ }+{30}^{\circ })+\sec ({900}^{\circ }+0^{\circ })$
$=-\sin ({810}^{\circ }+{60}^{\circ })-\text{cosec}({720}^{\circ }-{60}^{\circ })-\tan ({810}^{\circ }+{45}^{\circ })-2\tan \left({30}^{\circ }\right)-\sin \left({30}^{\circ }\right)-\sec \left(0^{\circ }\right)$
$=-\frac{1}{2}+\frac{2}{\sqrt{3}}+1-\frac{2}{\sqrt{3}}-\frac{1}{2}-1$
$=-1$

$\left(3\right)$
$\cos \theta =\frac{4}{5},\theta \in (\frac{3\pi }{2},2\pi )\Rightarrow \sin \theta =-\frac{3}{5}$
$\cos \alpha =\frac{3}{5},\alpha \in (0,\frac{\pi }{2})\Rightarrow \sin \alpha =\frac{4}{5}$

Now,
$\cos (\theta -\alpha )=\cos \theta \cos \alpha +\sin \theta \sin \alpha =\frac{4}{5}\times \frac{3}{5}-\frac{3}{5}\times \frac{4}{5}=0$