MATHS M.C.Q (1 Marks)

2. $-\frac{\text{b}}{\text{a}}$

Solution:$$

Given:

$\sin\alpha+\sin\beta=\text{a}...(\text{i})$

$\cos\alpha-\cos\beta=\text{b}...(\text{ii})$

Dividing (i) by (ii):

$\Rightarrow\ \frac{\sin\alpha+\sin\beta}{\cos\alpha-\cos\beta}=\frac{\text{a}}{\text{b}}$

$\Rightarrow\ \frac{2\sin\big(\frac{\alpha+\beta}{2}\big)\cos\big(\frac{\alpha-\beta}{2}\big)}{-2\sin\big(\frac{\alpha+\beta}{2}\big)\sin\big(\frac{\alpha-\beta}{2}\big)}=\frac{\text{a}}{\text{b}}$ $\Big[\because\ \sin\text{A}+\sin\text{B}=2\sin\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\text{ and }\cos\text{A}+\cos\text{B}$

$\Rightarrow\ \frac{\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha-\beta}{2}\Big)}{-\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha-\beta}{2}\Big)}=\frac{\text{a}}{\text{b}}$

$\Rightarrow\ \cot\Big(\frac{\alpha-\beta}{2}\Big)=-\frac{\text{a}}{\text{b}}$

$\Rightarrow\ \frac{1}{\cot\Big(\frac{\alpha-\beta}{2}\Big)}=\frac{1}{-\frac{\text{a}}{\text{b}}}$

$\Rightarrow\ \tan\Big(\frac{\alpha-\beta}{2}\Big)=-\frac{\text{b}}{\text{a}}$

1. $0$

Solution:$$

$\cos52^\circ+\cos68^\circ+\cos172^\circ$

$=\ 2\cos\Big(\frac{52^\circ+68^\circ}{2}\Big)\cos\Big(\frac{52^\circ-68^\circ}{2}\Big)+\cos172^\circ$ $\Big[\because\ \cos\text{A}+\cos\text{B}=2\cos\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$

$=\ 2\cos60^\circ\cos(-8^\circ)+\cos172^\circ$

$=\ 2\times\frac{1}{2}\cos8^\circ+\cos172^\circ$

$=\ \cos8^\circ+\cos172^\circ$

$=\ 2\cos\Big(\frac{8^\circ+172^\circ}{2}\Big)\cos\Big(\frac{8^\circ-172^\circ}{2}\Big)$

$=\ 2\cos90^\circ\cos82^\circ$

$=\ 0$

2. $-\frac{1}{2}$

Solution:$$

$\sin(60^\circ+18^\circ)-\sin(60^\circ-18^\circ)-(\sin66^\circ-\sin6^\circ)$

$=\ \sin60^\circ\cos18^\circ+\cos60^\circ\sin18^\circ-\sin60^\circ\cos18^\circ\\ \ \ \ +\cos60^\circ\sin18^\circ-(2\cos36^\circ\frac{1}{2})$

$=\ 2\times\frac{1}{2}\cos18^\circ-\cos36^\circ$

$=\ \frac{\sqrt5-1}{4}-\frac{\sqrt5+1}{4}$

$=\ \frac{-2}{4}$

$=\ \frac{-1}{2}$

4. $\frac{\pi}{4}$

Solution:

It is given that $\tan\alpha=\frac{\text{x}}{\text{x}+1}\text{ and }\tan\beta=\frac{1}{2\text{x}+1}.$

Now,

$\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$

$=\ \frac{\frac{\text{x}}{\text{x}+1}+\frac{1}{2\text{x}+1}}{1-\frac{\text{x}}{\text{x}+1}\times\frac{1}{2\text{x}+1}}$

$=\ \frac{\frac{\text{x}(2\text{x}+1)+\text{x}+1}{(\text{x}+1)(2\text{x}+1)}}{\frac{(\text{x+1})(2\text{x}+1)-\text{x}}{(\text{x}+1)(2\text{x}+1)}}$

$=\ \frac{2\text{x}^2+\text{x}+\text{x}+1}{2\text{x}^2+3\text{x}+1-1}$

$=\ \frac{2\text{x}^2+2\text{x}+1}{2\text{x}^2+2\text{x}+1}$

$=\ 1$

$\therefore\ \tan(\alpha+\beta)=1=\tan\frac{\pi}{4}$

$\ \Rightarrow\ \alpha+\beta=\frac{\pi}{4}$

2. $0$

Solution:

We have,

$\sin\text{x}+\sin\text{y}=\sqrt3(\cos\text{y}-\cos\text{x})$

$\Rightarrow\ 2\sin\Big(\frac{\text{x+y}}{2}\Big)\cos\Big(\frac{\text{x}-\text{y}}{2}\Big)=2\sqrt3\sin\Big(\frac{\text{x+y}}{2}\Big)\sin\Big(\frac{\text{x}-\text{y}}{2}\Big)$

$\Rightarrow\ 2\sin\Big(\frac{\text{x+y}}{2}\Big)\cos\Big(\frac{\text{x}-\text{y}}{2}\Big)-2\sqrt3\sin\Big(\frac{\text{x+y}}{2}\Big)\sin\Big(\frac{\text{x}-\text{y}}{2}\Big)=0$

$\Rightarrow\ 2\sin\Big(\frac{\text{x+y}}{2}\Big)\Big[\cos\Big(\frac{\text{x}-\text{y}}{2}\Big)-\sqrt3\sin\Big(\frac{\text{x}-\text{y}}{2}\Big)\Big]=0$

$\Rightarrow\ \sin\Big(\frac{\text{x+y}}{2}\Big)\Big[\cos\Big(\frac{\text{x}-\text{y}}{2}\Big)-\sqrt3\sin\Big(\frac{\text{x}-\text{y}}{2}\Big)\Big]=0$

$\Rightarrow\ \sin\frac{\text{x+y}}{2}=0$ Or, $\cos\Big(\frac{\text{x}-\text{y}}{2}\Big)-\sqrt3\sin\Big(\frac{\text{x}-\text{y}}{2}\Big)=0$

$\Rightarrow\ \frac{\text{x+y}}{2}=0$ Or, $\tan\Big(\frac{\text{x}-\text{y}}{2}\Big)=\frac{1}{\sqrt3}=\tan\frac{\pi}{6}$

$\Rightarrow\ \text{x}=-\text{y}\text{ Or }, \frac{\text{x}-\text{y}}{2}=\frac{\pi}{6}$

$\Rightarrow\ \text{x}=-\text{y}\text{ Or }, \text{x}-\text{y}=\frac{\pi}{3}$

Case 1:

When $\text{x}=-\text{y}$

In this case,

$\sin3\text{x}+\sin3\text{y}=\sin(-3\text{y})+\sin3\text{y}=-\sin3\text{y}+\sin3\text{y}=0$

Case 2:

When $\text{x}-\text{y}=\frac{\pi}{3}$

Or, $3\text{x}=\pi+3\text{y}$

So, $\sin3\text{x}+\sin3\text{y}=\sin(\pi+3\text{y})+\sin3\text{y}$

$=\ -\sin3\text{y}+\sin3\text{y}$

$=\ 0$

2. $0$

Solution:$$

$\sin50^\circ-\sin70^\circ+\sin10^\circ$

$=\ 2\sin\Big(\frac{50^\circ-70^\circ}{2}\Big)\cos\Big(\frac{50^\circ+70^\circ}{2}\Big)+\sin10^\circ$ $\Big[\because\ \sin\text{A}-\sin\text{B}=2\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\Big]$

$=\ 2\sin(-10^\circ)\cos60^\circ+\sin10^\circ$

$=\ 2\times\frac{1}{2}\sin(-10^\circ)+\sin10^\circ$

$=\ -\sin10^\circ+\sin10^\circ$

$=\ 0$

2. $\text{HP}$

Solution:

 Given:

$\sin(\text{B+C}-\text{A}),\sin(\text{C+A}-\text{B}),\text{ and }\sin(\text{A+B}-\text{C})$ are in A.P.

$\Rightarrow\ \sin(\text{C+A}-\text{B})-\sin(\text{B+C}-\text{A})\\ \ \ \ =\sin(\text{A+B}-\text{C})-\sin(\text{C+A}-\text{B})$

$\Rightarrow\ 2\sin\Big(\frac{\text{C+A}-\text{B}-\text{B}-\text{C+A}}{2}\Big)\cos\Big(\frac{\text{C+A}-\text{B+B+C}-\text{A}}{2}\Big)\\ \ \ \ =2\sin\Big(\frac{\text{A+B}-\text{C}-\text{C}-\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A+B}-\text{C+C+A}-\text{B}}{2}\Big)$

$\Rightarrow\ \sin(\text{A}-\text{B})\cos\text{C}=\sin(\text{B}-\text{C})\cos\text{A}$

$\Rightarrow\ \sin\text{A}\cos\text{B}\cos\text{C}-\cos\text{A}\sin\text{B}\cos\text{C}\\ \ \ =\sin\text{B}\cos\text{C}\cos\text{A}-\cos\text{B}\sin\text{C}\cos\text{A}$

$\Rightarrow\ 2\sin\text{B}\cos\text{A}\cos\text{C}=\sin\text{A}\cos\text{B}\cos\text{C}+\cos\text{A}\cos\text{B}\sin\text{C}$

Dividing both sides by $\cos\text{A}\cos\text{B}\cos\text{C}:$

$2\tan\text{B}=\tan\text{A}+\tan\text{C}$

$\Rightarrow\ \frac{2}{\cot\text{B}}=\frac{1}{\cot\text{A}}+\frac{1}{\cot\text{C}}$

Hence, $\cot\text{A}, \cot\text{B}\text{ and }\cot\text{C}$ are in H.P. 

2. $\frac{5}{8}$

Solution:$$

Given:

$\sin2\theta+\sin2\phi=\frac{1}{2}....(\text{i})$

and

$\cos2\theta+\cos2\phi=\frac{3}{2}....(\text{ii})$

Squaring and adding (i) and (ii), we get:

$(\sin2\theta+\sin2\phi)^2+(\cos2\theta+\cos2\phi)^2=\frac{1}{4}+\frac{9}{4}$

$\Rightarrow\ \Big[2\sin\Big(\frac{2\theta+2\phi}{2}\Big)\cos\Big(\frac{2\theta-2\phi}{2}\Big)\Big]^2\\ \ \ \ \ +\Big[2\cos\Big(\frac{2\theta+2\phi}{2}\Big)\cos\Big(\frac{2\theta-2\phi}{2}\Big)\Big]^2=\frac{5}{2}$

$\Rightarrow\ 4\sin^2(\theta+\phi)\cos^2(\theta-\phi)\\ \ \ +4\cos^2(\theta+\phi)\cos^2(\theta-\phi)=\frac{5}{2}$

$\Rightarrow\ 4\cos^2(\theta-\phi)[\sin^2(\theta+\phi)+\cos^2(\theta+\phi)]=\frac{5}{2}$

$\Rightarrow\ 4\cos^2(\theta-\phi)=\frac{5}{2}$

$\Rightarrow\ \cos^2(\theta-\phi)=\frac{5}{8}$

3. $\frac{\text{m}+1}{\text{m-1}}$

Solution:

Given:

$\cos\text{A}=\text{m}\cos\text{B}$

$\Rightarrow\ \frac{\cos\text{A}}{\cos\text{B}}=\frac{\text{m}}{1}$

$\Rightarrow\ \frac{\cos\text{A}+\cos\text{B}}{\cos\text{A}-\cos\text{B}}=\frac{\text{m}+1}{\text{m}-1}$

$\Rightarrow\ \frac{2\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A+B}}{2}\Big)}{-2\sin\Big(\frac{\text{B+A}}{2}\Big)\sin\Big(\frac{\text{B}-\text{A}}{2}\Big)}=\frac{\text{m}+1}{\text{m}-1}$ $\Big[\because\ \cos\text{A}+\cos\text{B}=2\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A+B}}{2}\Big)\\\text{ and } \cos\text{A}-\cos\text{B}=2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$

$\Rightarrow\ \frac{\cos\Big(\frac{\text{B}-\text{A}}{2}\Big)\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)}{\sin\Big(\frac{\text{A+B}}{2}\Big)\sin\Big(\frac{\text{B}-\text{A}}{2}\Big)}=\frac{\text{m}+1}{\text{m}-1}$

$\Rightarrow\ \cot\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{B}-\text{A}}{2}\Big)=\frac{\text{m}+1}{\text{m}-1}$

4. $-\frac{1}{2}$

Solution:$$

$\cos40^\circ+\cos80^\circ+\cos160^\circ+\cos240^\circ$

$=\ 2\cos\Big(\frac{40^\circ+80^\circ}{2}\Big)\cos\Big(\frac{40^\circ-80^\circ}{2}\Big)\\ \ \ \ +\cos160^\circ-\cos(180^\circ+60^\circ)$$\big[\because\ \cos\text{A}+\cos\text{B}=2\big]$

$=\ 2\cos60^\circ\cos(-20^\circ)+\cos160^\circ-\frac{1}{2}$

$=\ 2\times\frac{1}{2}\cos20^\circ+\cos160^\circ-\frac{1}{2}$

$=\ -\cos(180-20)^\circ+\cos160^\circ-\frac{1}{2}$

$=\ -\frac{1}{2}$

4. $\cos7^\circ$

Solution:$$

$\sin47^\circ+\sin61^\circ-\sin11^\circ-\sin25^\circ$

$=\ \sin47^\circ-\sin25^\circ+\sin61^\circ-\sin11^\circ$

$=\ 2\sin\Big(\frac{47^\circ-25^\circ}{2}\Big)\cos\Big(\frac{47^\circ+25^\circ}{2}\Big)\\ \ \ \ +2\sin\Big(\frac{61^\circ-11^\circ}{2}\Big)\cos\Big(\frac{61^\circ+11^\circ}{2}\Big)$

$=\ 2\sin11^\circ\cos36^\circ+2\sin25^\circ\cos36^\circ$

$=\ 2\cos36^\circ(\sin11^\circ+\sin25^\circ)$

$=\ 2\cos36^\circ\Big\{2\sin\Big(\frac{11^\circ+25^\circ}{2}\Big)\cos\Big(\frac{11^\circ-25^\circ}{2}\Big)\Big\}$

$=\ 4\cos36^\circ\sin18^\circ\cos7^\circ$

$=\ 4\times\Big(\frac{\sqrt5-1}{4}\Big)\Big(\frac{\sqrt5+1}{4}\Big)\cos7^\circ$ $\Big[\cos36^\circ=\frac{\sqrt5+1}{4}\text{ and }\sin18^\circ=\frac{\sqrt5-1}{4}\Big]$

$=\ \frac{5-1}{4}\cos7^\circ$

$=\ \cos7^\circ$

2. $\frac{1}{2}$

Solution:$$

$\sin163^\circ\cos347^\circ+\sin73^\circ\sin167^\circ$

$=\ \sin(180^\circ-17^\circ)\cos(360^\circ-13^\circ)\\ \ \ +\sin(90^\circ-17^\circ)\sin(180^\circ+13^\circ)$

$=\ \sin17^\circ\cos13^\circ+\cos17^\circ\sin13^\circ$

$=\ \sin(17^\circ+13^\circ)$ $[\sin(\text{A+B})=\sin\text{A}\cos\text{B}+\sin\text{B}\cos\text{A}]$

$=\ \sin30^\circ$

$=\ \frac{1}{2}$

1. $0$

Solution:

$\cos35^\circ+\cos85^\circ+\cos155^\circ$

$=\ 2\cos\Big(\frac{35^\circ+85^\circ}{2}\Big)\cos\Big(\frac{35^\circ-85^\circ}{2}\Big)+\cos155^\circ$ $\Big[\because\ \cos\text{A}+\cos\text{B}=2\cos\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$

$=\ 2\cos60^\circ\cos(-25^\circ)+\cos155^\circ$

$=\ 2\times\frac{1}{2}\cos25^\circ+\cos155^\circ$

$=\ \cos25^\circ+\cos155^\circ$

$=\ 2\cos\Big(\frac{25^\circ+155^\circ}{2}\Big)\cos\Big(\frac{25^\circ-155^\circ}{2}\Big)$

$=\ 2\cos90^\circ\cos65^\circ$

$=\ 0$

2. $\cot\text{B}$

Solution:

Since A,B and C are in A.P,

$\text{B}-\text{A}=\text{C}-\text{B}$

$\text{Or },2\text{B}=\text{A}+\text{C}$

$\frac{\sin\text{A}-\sin\text{C}}{\cos\text{C}-\cos\text{A}}$

$=\ \frac{2\sin\Big(\frac{\text{A}-\text{C}}{2}\Big)\cos\Big(\frac{\text{A+C}}{2}\Big)}{-2\sin\Big(\frac{\text{C+A}}{2}\Big)\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)}$ $\Big[\because\ \sin\text{A}-\sin\text{B}=2\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A+B}}{2}\Big)\\\text{ and }\cos\text{A}-\cos\text{B}=-2\sin\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$

$=\ \frac{\sin\Big(\frac{\text{A}-\text{C}}{2}\Big)\cos\Big(\frac{\text{A+C}}{2}\Big)}{-\sin\Big(\frac{\text{A+C}}{2}\Big)\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)}$

$=\ \frac{\sin\Big(\frac{\text{A}-\text{C}}{2}\Big)\cos\Big(\frac{\text{A+C}}{2}\Big)}{\sin\Big(\frac{\text{A+C}}{2}\Big)\sin\Big(\frac{\text{C}-\text{A}}{2}\Big)}$

$=\ \frac{\cos\Big(\frac{\text{A+C}}{2}\Big)}{\sin\Big(\frac{\text{A+C}}{2}\Big)}$

$=\ \frac{\cos\text{B}}{\sin\text{B}}$

$=\ \cot\text{B}$