MATHS M.C.Q (1 Marks)

4. $\text{f(x)}\geq2$

Solution:

Given that; $\text{f(x)}=\cos^2\text{x}+\sec^2\text{x}$

We know that $\text{AM}\geq\text{GM}$

$\Rightarrow\frac{\cos^2\text{x}+\sec^2\text{x}}{2}\geq\sqrt{\cos^2\text{x}.\sec^2\text{x}}$

$\Rightarrow\frac{\cos^2\text{x}+\sec^2\text{x}}{2}\geq1$ $\big[\text{since}\sec\theta=\frac{1}{\cos\theta}\big]$

$\Rightarrow\cos^2\text{x}+\sec^2\text{x}\geq2$

$\Rightarrow\text{f(x)}\geq2$

Hence, the correct option is (d)

3. $\frac{44}{117}$

Solution:

We have:

$\text{cosec}\text{ x}+\cot\text{x}=\frac{11}{2}\cdots(1)$

$\Rightarrow\frac{1}{\text{cosec}\text{ x}+\cot\text{x}}=\frac{2}{11}$

$\Rightarrow\frac{\text{cosec}^2\text{x}-\cot\text{x}}{\text{cosec }\text{x}-\cot\text{x}}=\frac{2}{11}$

$\Rightarrow\frac{(\text{cosec}\text{ x}+\cot\text{x})(\text{cosec}\text{ x}-\cot\text{x})}{(\text{cosec}\text{ x}+\cot\text{x})}=\frac{2}{11}$

$\therefore\text{cosec}\text{ x}-\cot\text{x}=\frac{2}{11}\cdots(2)$

subtracting (2) from (1):

$\Rightarrow 2\cot\text{x}=\frac{11}{2}-\frac{2}{11}$

$\Rightarrow2\cot\text{x}=\frac{121-4}{22}$

$\Rightarrow2\cot\text{x}=\frac{117}{22}$

$\Rightarrow\cot\text{x}=\frac{117}{44}$

$\Rightarrow\frac{1}{\tan\text{x}}=\frac{117}{44}$

$\Rightarrow\tan\text{x}=\frac{44}{117}$

3. $\tan\frac{\alpha}{2}$

Solution:

$\frac{\sin5\alpha-\sin3\alpha}{\cos5\alpha+2\cos4\alpha+\cos3\alpha}=\frac{\sin5\alpha-\sin3\alpha}{\cos5\alpha+\cos3\alpha+2\cos4\alpha}$

$=\frac{2\sin\alpha\cos4\alpha}{2\cos4\alpha\cos\alpha+2\cos4\alpha}$

$=\frac{2\sin\alpha\cos4\alpha}{2\cos4\alpha(\cos\alpha+1)}$

$=\frac{\sin\alpha}{\cos\alpha+1}$

$=\frac{2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}+\sin^2\frac{\alpha}{2}+\cos^2\frac{\alpha}{2}}$

$=\frac{2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2\cos^2\frac{\alpha}{2}}$

$=\frac{\sin\frac{\alpha}{2}}{\cos^2\frac{\alpha}{2}}$

$=\tan\frac{\alpha}{2}$

3. $\frac16$

Solution:.

In $\triangle\text{ABC},$

$\text{A+B+C}=\pi$

We know that $\tan(\text{A+B+C)}=\frac{\tan\text{A}+\tan\text{B}+\tan\text{C}-\tan\text{A}\tan\text{B}\tan\text{C}}{1-\tan\text{A}\tan\text{B}-\tan\text{B}\tan\text{C}-\tan\text{C}\tan\text{A}}$ and $\tan\pi=0.$

$\therefore\tan\text{A}+\tan\text{B}+\tan\text{C}-\tan\text{A}\tan\text{B}\tan\text{C}=0$

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

If $\tan\text{A}+\tan\text{B}+\tan\text{C}=6,\tan\text{A}\tan\text{B}\tan\text{C}=6$

$\Rightarrow\frac{1}{\tan\text{A}\tan\text{B}\tan\text{C}}=\frac16$

$\Rightarrow\cot\text{A}\cot\text{B}\cot\text{C}=\frac16$

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):

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

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

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

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

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

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

3. 2

Solution:

We have:

$\sin^2\frac{\pi}{18}+\sin^2\frac{\pi}{9}+\sin^2\frac{7\pi}{2}+\sin^2\frac{4\pi}{9}$

$=\sin^2\frac{\pi}{18}+\sin^2\frac{2\pi}{8}+\sin^2\frac{7\pi}{18}+\sin^2\frac{8\pi}{2}$

$=\sin^2\frac{\pi}{18}+\sin^2\frac{2\pi}{8}+\sin^2\Big(\frac{7\pi}{18}\Big)+\sin^2\Big(\frac{8\pi}{2}\Big)$

$=\sin^2\frac{\pi}{18}+\sin^2\frac{2\pi}{18}+\sin^2\Big(\frac{\pi}{2}-\frac{\pi}{18}\Big)+\sin^2\Big(\frac{\pi}{2}-\frac{\pi}{18}\Big)$

$=\sin^2\frac{\pi}{18}+\sin^2\frac{2\pi}{18}+\cos^2\frac{2\pi}{18}+\cos^2\frac{\pi}{18}$

$=\sin^2\frac{\pi}{18}+\sin^2\frac{\pi}{18}+\cos^2\frac{2\pi}{18}+\cos^2\frac{2\pi}{18}$

$=1+1$

$=2$

4. $\frac{-3}{130}$

2. $\sqrt{6}$

Solution:

It is given that $\text{a}=2,\angle\text{B}=60^{\circ}$ and $\angle\text{C}=75^{\circ}.$

In $\triangle\text{ABC},$

$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^{\circ}$ (Angle sum property)

$\Rightarrow\angle\text{A}+60^{\circ}+75^{\circ}=180^{\circ}$

$\Rightarrow\angle\text{A}=180^{\circ}-135^{\circ}=45^{\circ}$

Using sine rule, we get

$\frac{2}{\sin45^{\circ}}=\frac{\text{b}}{\sin60^{\circ}}$ $\Big(\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}\Big)$

$\Rightarrow\text{b}=\frac{2\times\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}}=\sqrt{6}$

Hence, the correct answer is option (b).

2. $\frac{23}{10}$

Solution:

It is given that $\frac{\pi}{2}<\text{A}<\pi$

$3\tan\text{A}+4=0$

$\Rightarrow\tan\text{A}=-\frac{4}{3}$

$\Rightarrow\cot\text{A}=-\frac{3}{4}$

Now, 

$\sec\text{A}=\pm\sqrt{1+\tan^2\text{A}}$

$=\pm\sqrt{1+\frac{16}{9}}$

$=\pm\sqrt{\frac{25}{9}}=\pm\frac{5}{3}$

Also,

$\sin\text{A}=\pm\sqrt{1-\cos^2\text{A}}$

$=\pm\sqrt{1-\frac{9}{25}}$

$=\pm\sqrt{\frac{16}{25}}=\pm\frac{4}{5}$

$\therefore\sin\text{A}=\frac{4}{5}$ (A lines in 2 nd quadrant)

so,

$2\cot\text{A}-5\cos\text{A}+\sin\text{A}$

$=2\times\Big(-\frac{3}{5}\Big)-5\times\Big(-\frac{3}{5}\Big)+\frac{4}{5}$

$=-\frac{3}{2}+3+\frac{4}{5}$

$=-\frac{15+30+8}{10} $

$=\frac{23}{10}$

Hence,the correct answer is option B.

1. $\frac{1+\text{k}}{1-\text{k}}$

Solution:

$\frac{\cos(\theta_1-\theta_2)}{\cos(\theta_1+\theta_2)}$

$=\frac{\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2}{\cos\theta_1\cos\theta_2-0\sin\theta_1\sin\theta_2}$

Dividing numerator and denominator by $\cos\theta_1\cos\theta_2,$ we get:

$\frac{1+\tan\theta_1\tan\theta_2}{1-\tan\theta_1\tan\theta_2}$

$=\frac{1+\text{k}}{1-\text{k}}$

3. $\frac{44}{117}$

Solution:

We have:

$\text{cosec}\text{ x}+\cot\text{x}=\frac{11}{2}\cdots(1)$$$

$\Rightarrow\frac{1}{\text{cosec}\text{ x}+\cot\text{x}}=\frac{2}{11}$

$\Rightarrow\frac{\text{cosec}^2\text{x}-\cot\text{x}}{\text{cosec}\text{ x}-\cot\text{x}}=\frac{2}{11}$

$\Rightarrow\frac{(\text{cosec}\text{ x}+\cot\text{x})(\text{cosec}\text{ x}-\cot\text{x})}{(\text{cosec}\text{ x}+\cot\text{x})}=\frac{2}{11}$

$\therefore\text{cosec}\text{ x}-\cot\text{x}=\frac{2}{11}\cdots(2)$

subtracting (2) from (1):

$\Rightarrow 2\cot\text{x}=\frac{11}{2}-\frac{2}{11}$

$\Rightarrow2\cot\text{x}=\frac{121-4}{22}$

$\Rightarrow2\cot\text{x}=\frac{117}{22}$

$\Rightarrow\cot\text{x}=\frac{117}{44}$

$\Rightarrow\frac{1}{\tan\text{x}}=\frac{117}{44}$

$\Rightarrow\tan\text{x}=\frac{44}{117}$

3.  $\text{x}=\frac{\text{n}\pi}{7}+\frac{\pi}{14},\ \text{n}\in\text{Z}$

Solution:

Given:

$\tan5\text{x}=\cot2\text{x}$

$\Rightarrow\tan5​\text{x}=\tan\Big(\frac{\pi}{2}-2\text{x}\Big)$

$\Rightarrow5\text{x}=\text{n}\pi+\frac{\pi}{2}-2\text{x}$

$\Rightarrow7\text{x}=\text{n}\pi+\frac{\pi}{2}$

$\Rightarrow\text{x}=\frac{\text{n}\pi}{7}+\frac{\pi}{14},\ \text{n}\in\text{Z}$ 

2. 0

Solution:

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

$=2\cos\Big(\frac{50^\circ+70^\circ}{2}\Big)\sin\Big(\frac{50^\circ-70^\circ}{2}\Big)+\sin10^\circ$

$=-2\cos60^\circ\sin10^\circ+\sin10^\circ=-2\cdot\frac{1}{2}\sin10^\circ+\sin10^\circ=0$ 

1. $\text{c}^2-3\text{c}-7=0$

Solution:

It is given that a = 4, b = 3 and $\angle\text{A}=60^{\circ}$

Using cosine rule, we have

$\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}$

$\Rightarrow\cos60^{\circ}=\frac{9+\text{c}^2-16}{2\times3\times\text{c}}$

$\Rightarrow\frac{1}{2}=\frac{\text{c}^2-7}{6\text{c}}$

$\Rightarrow\text{c}^2-7=3\text{c}$

$\Rightarrow\text{c}^2-3\text{c}-7=0$

Thus, c is the root of $\text{c}^2-3\text{c}-7=0.$

Hence, the correct answer is option (a).

1. $3+\frac{2\sqrt{2}}{2}$

4. 4

Solution:

We have,

$\big(\cot\frac{\text{x}}{2}-\tan\frac{\text{x}}{2}\big)(1-2\tan\text{x}\cot2\text{x})$

$\big(\cot^2\frac{\text{x}}{2}-2\cot\frac{\text{x}}{2}+\tan^2\frac{\text{x}}{2}\big)\Big\{1-2\tan\text{x}\Big(\frac{\cot^2\text{x}-1}{2\cot\text{x}}\Big)\Big\}$

$\big(\cot^2\frac{\text{x}}{2}-2+\tan^2\frac{\text{x}}{2}\big)\Big\{1-\tan\text{x}\Big(\frac{\cot^2\text{x}-1}{\cot\text{x}}\Big)\Big\}$

$\big(\cot^2\frac{\text{x}}{2}+\tan^2\frac{\text{x}}{2}-2\big)\Big(1-\frac{\cot^2\text{x}-\tan\text{x}}{\cot\text{x}}\Big)$

$\big(\cot^2\frac{\text{x}}{2}+\tan^2\frac{\text{x}}{2}-2\big)\big(\tan^2\text{x}\big)$

$\big(\cot^2\frac{\text{x}}{2}+\tan\frac{\text{x}}{2}-2\big)\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1-\tan^2\frac{\text{x}}{2}}\Bigg)^2$

$=\frac{1}{\big(1-\tan^2\frac{\text{x}}{2}\big)^2}\big(4+4\tan^4\frac{\text{x}}{2}-8\tan^2\frac{\text{x}}{2}\big)$

$=\frac{1}{\big(1-\tan^2\frac{\text{x}}{2}\big)^2}\big(4-8\tan^2\frac{\text{x}}{2}+4\tan^4\frac{\text{x}}{2}\big)$

$=\frac{1}{\big(1-\tan^2\frac{\text{x}}{2}\big)^2}\Big\{\big(\tan^2\frac{\text{x}}{2}\big)^2-2\big(\tan^2\frac{\text{x}}{2}+1\big)\Big\}$

$=\frac{4\big(\tan^2\frac{​​\text{x}}{2}-1\big)^2}{\big(1-\tan^2\frac{\text{x}}{2}\big)^2}$

$=4$

3. $\frac{-3}{\sqrt{10}}$

Solution:

$\tan\theta=3,\theta$ lies in third quadrant, it is positive.

$\tan\theta=\frac{\text{P}}{\text{B}}=\frac{3}{1}$ 

Then, Hypotenuse $=\sqrt{(3)^2+(1)^2}=\sqrt{9+1}=\sqrt10$

$\therefore\sin\theta=\frac{3}{\sqrt{10}}$ where $\theta$ lies in third quadrant

Hence the correct option is (c).

2. $\text{x=y, x}\neq0$

Solution:

We have:

$\sec^2\text{x}=\frac{4\text{xy}}{(\text{x}+\text{y})^2}$

$\Rightarrow\frac{4\text{x}\text{y}}{(\text{x}+\text{y})^2}\geq1 $ $[\therefore\sec^2\text{x}\geq1]$

$\Rightarrow4\text{xy}\geq(\text{x}+\text{y})^2$

$\Rightarrow4\text{xy}\geq\text{x}^2+\text{y}^2+2\text{xy}$

$\Rightarrow2\text{xy}\geq\text{x}^2+\text{y}^2$

$\Rightarrow(\text{x}-\text{y})^2\leq0$

$\Rightarrow(\text{x}-\text{y})\leq0$

$\Rightarrow\text{x}=\text{y}$

For $\text{x}=0,\sec^2\text{x}$ will not be defined,

$\Rightarrow\text{x}\neq 0$

$\therefore\text{x}=\text{y}$

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:$$

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

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

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

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

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

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

2. $\text{b}$

Solution:

Givan: $\tan\text{x}=\frac{​​\text{a}}{\text{b}}$

Now,

$=\cos2\text{x}+\alpha\sin2\text{x}$

$=\text{b}\Big(\frac{1-\tan^2\text{x}}{1+\tan^2\text{x}}\Big)+\text{a}\Big(\frac{2\tan\text{x}}{1+\tan^2\text{x}}\Big)$

$=\text{b}\Bigg(\frac{1-\frac{\text{a}^2}{\text{b}^2}}{1+\frac{\text{a}^2}{\text{b}^2}}\Bigg)+\text{a}\Bigg(\frac{1\times\frac{\text{a}}{\text{b}}}{1+\frac{\text{a}^2}{\text{b}^2}}\Bigg)$

$=\frac{\text{b}(\text{b}^2-\text{a}^2)}{\text{a}^2+\text{b}^2}+\frac{2\text{a}^2\text{b}^2}{\text{a}^2+\text{b}^2}$

$=\frac{\text{b}^3-\text{a}^2\text{b}+2\text{a}^2\text{b}}{\text{a}^2+\text{b}^2}$

$=\frac{\text{b}^3+\text{a}^2\text{b}}{\text{a}^2+\text{b}^2}$

$=\frac{\text{b}(\text{b}^2+\text{a}^2)}{\text{a}^2+\text{b}^2}$

$=\text{b}$

3. $\text{a}^2+\text{b}^2+\text{c}^2$

Solution:

Using cosine rule, we have

$2(\text{bc}\cos\text{A}+\text{ca}\cos\text{B}+\text{ab}\cos\text{C})$

$=2\text{bc}\Big(\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}\Big)+2\text{ca}\Big(\frac{\text{c}^2+\text{a}^2-\text{b}^2}{2\text{ca}}\Big)+2\text{ab}\Big(\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}\Big)$

$=\text{b}^2+\text{c}^2-\text{a}^2+\text{c}^2+\text{a}^2-\text{b}^2+\text{a}^2+\text{b}^2-\text{c}^2$

$=\text{a}^2+\text{b}^2+\text{c}^2$

Hence, the correct answer is option (c). 

4. $0$

Solution:

Using sine rule, we have

$\sum\text{a}^2(\sin\text{B}-\sin\text{C})$

$\text{a}^2\Big(\frac{\text{b}}{\text{k}}-\frac{\text{c}}{\text{k}}\Big)+\text{b}^2\Big(\frac{\text{c}}{\text{k}}-\frac{\text{a}}{\text{k}}\Big)+\text{c}^2\Big(\frac{\text{a}}{\text{k}}-\frac{\text{b}}{\text{k}}\Big)$

$=\frac{1}{\text{k}}(\text{a}^2\text{b}-\text{a}^2\text{c}+\text{b}^2\text{c}-\text{b}^2\text{a}+\text{c}^2\text{a}-\text{c}^2\text{b})$

This expression cannot be simplified to match with any of the given options.

However, if the quesion is "In any $\triangle\text{ABC},\sum\text{a}^2(\sin\text{B}-\sin\text{C})=",$ then the solution is as follows.

Using sine rule, we have

$\sum\text{a}^2(\sin^2\text{B}-\sin^2\text{C})$

$\text{a}^2\Big(\frac{\text{b}^2}{\text{k}^2}-\frac{\text{c}^2}{\text{k}^2}\Big)+\text{b}^2\Big(\frac{\text{c}^2}{\text{k}^2}-\frac{\text{a}^2}{\text{k}^2}\Big)+\text{c}^2\Big(\frac{\text{a}^2}{\text{k}^2}-\frac{\text{b}^2}{\text{k}^2}\Big)$

$\frac{1}{\text{k}^2}(\text{a}^2\text{b}^2-\text{a}^2\text{c}^2+\text{b}^2\text{c}^2-\text{b}^2\text{a}^2+\text{c}^2\text{a}^2-\text{c}^2\text{b}^2)$

$=\frac{1}{\text{k}^2}\times0$

$=0$

Hence, the correct answer is option (d).

1. $1$

Solution:

We have,
$\Rightarrow2\cos^2​​\text{x}-1+2\cos\text{x}=1$

$\Rightarrow\cos^2\text{x}+\cos\text{x}-1=0$

$\Rightarrow\cos\text{x}=\frac{-1\pm\sqrt{1^2+4}}{2}$

$\Rightarrow\cos\text{x}=\frac{-1\pm\sqrt{5}}{2}$

$\Rightarrow\cos\text{x}=\frac{-1+\sqrt{5}}{2}$

Now,

$(2-\cos^2​​\text{x})\sin^2\text{x}=\bigg[2-\Big(\frac{-1+\sqrt{5}}{2}\Big)^2\bigg](1-\cos^2\text{x})$

$=\bigg[2-\frac{1}{4}\Big(1-2\sqrt{5}+5\Big)\bigg]\Big(1-\frac{1}{4}\Big(1-2\sqrt{5}+5\Big)\Big)$

$=\frac{1}{4}\Big(1+\sqrt{5}\Big)\Big(\sqrt{5}-1\Big)=\frac{4}{4}=1$

2. $\tan\frac{\text{x}}{2}$

Solution:

We have:

$\frac{\text{y}+1}{1-\text{y}}=\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}$

$\Rightarrow\frac{\text{y}+1}{1-\text{y}}=\sqrt{\frac{\cos^2\frac{\text{x}}{2}+\sin^2\frac{\text{x}}{2} +2\sin\frac{\text{x}}{2}-\cos\frac{\text{x}}{2}}{\cos^2\frac{\text{x}}{2}+\sin^2\frac{\text{x}}{2}-2\sin\frac{\text{x}}{2}\cos\frac{\text{x}}{2}}}$

$\Rightarrow\frac{\text{y}+1}{1-\text{y}}=\sqrt{\frac{\Big(cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\Big)^2}{\Big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}\Big)^2}}$

$\Rightarrow\frac{\text{y}+1}{1-\text{y}}=\frac{\Big(\cos\frac{\text{x}}{2}+\sin\frac{\text{x}}{2}\Big)}{\Big(\cos\frac{\text{x}}{2}-\sin\frac{\text{x}}{2}\Big)}$ $[\because 0< \text{x }<\frac{\pi}{2}\Rightarrow0< \frac{\pi}{2}<\frac{\pi}{4}, 0 \text{ to } \frac{\pi}{4}\cos\text{x}>\sin\text{x}]$

$\Rightarrow\frac{\frac{\cos\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}} +\frac{\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}}}{\frac{\cos\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}}-\frac{\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}}}$

$\Rightarrow\frac{\text{y}+1}{1-\text{y}}=\frac{1+\tan\frac{\text{x}}{2}}{1-\tan\frac{\text{x}}{2}}$

comparing both the sides: 

$\text{y}=\tan\frac{\text{x}}{2}$

2. 0

Solution:

$\cos1^\circ \cos2^\circ \cos3^\circ...\ \cos179^\circ$

$=\cos1^\circ\cos2^\circ\cos3^\circ...\ \cos90^\circ...\ \cos179^\circ$

$=0$ $(\cos90^\circ = 0)$

Hence, the correct answer is option B.

2. $\frac{23}{10}$

Solution:

Given that, $3\tan\text{A}+4=0,$ A lies in second quadrant 

$\therefore\tan\text{A}=\frac{-4}{3}$ 

$\cos\text{A}=\frac{-3}{5}$ [A lies in second quadrant]

and $\sin\text{A}=\frac{4}{5}$ and $\cot\text{A}=\frac{-3}{4}$

$\therefore2\cot\text{A}-5\cos\text{A}+\sin\text{A}=2\Big(\frac{-3}{4}\Big)-5\Big(\frac{-3}{5}\Big)+\frac{4}{5}\\=\frac{-3}{2}+3+\frac{4}{5}=\frac{-15+30+8}{10}=\frac{23}{10}$ 

Hence, the correct option is (b).

2. 1

Solution:

$\tan1^\circ\tan2^\circ\tan^\circ\dots\tan89^\circ$

$=[\tan1^\circ\tan2^\circ\dots\tan44^\circ]\\\tan45^\circ[\tan(90^\circ-44^\circ)\tan(90^\circ-43^\circ)\dots\tan(90^\circ-1^\circ)]$

$=[\tan1^\circ\tan2^\circ\dots\tan44^\circ][\cot44^\circ\cot43^\circ\dots\cot1^\circ]$

$=1-1\dots1-1=1$

2. $\text{x}=\text{n}\pi\pm\frac{\pi}{3},\ \text{n}\in\text{Z}$

Solution:

Given:

$\sqrt{3}\sin​\text{x}+\cos\text{x}=\sqrt{3}\ .....(1)$

This equation is of the form $\text{a}\sin\theta+\text{b}\cos\theta=\text{c},$ where $\text{a}=\sqrt{3},\ \text{b}=1$ and $\text{c}=\sqrt{3}$

Let:

$\text{a}=\text{r}\cos\alpha$ and $\text{b}=\text{r}\sin\alpha$

Now

$\text{r}=\sqrt{\text{a}^2+\text{b}^2}=\sqrt{\Big(\sqrt{3}\Big)^2+1^1}=2$ and $\tan\alpha=\frac{\text{b}}{\text{a}}\Rightarrow\tan\alpha=\frac{1}{\sqrt{3}}$

$\Rightarrow\alpha=\frac{\pi}{6}$

On putting $\text{a}=\sqrt{3}=\text{r}\cos\alpha$ and $\text{b}=1=\text{r}\sin\alpha$ in equation (1),  we get:

$​\text{r}​\cos\alpha\sin​\text{x}​+​\text{r}​\sin\alpha\cos​\text{}x​=\sqrt{3}$

$\Rightarrow\text{r}\sin\big(\text{x}+\alpha\big)=\sqrt{3}$

$\Rightarrow\text{2}\sin\big(\text{x}+\alpha\big)=\sqrt{3}$

$\Rightarrow\text{}\sin\big(\text{x}+\alpha\big)=\frac{\sqrt{3}}{2}$

$\Rightarrow\text{}\sin\big(\text{x}+\alpha\big)=\sin\frac{\pi}{3}$

$\Rightarrow\text{}\sin\big(\text{x}+\frac{\pi}{6}\big)=\sin\frac{\pi}{3}$

$\Rightarrow​​\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{3}-\frac{\pi}{6},\ \text{n}\in\text{Z}$

1. $\frac{1}{2}\cos2\text{x}$

Solution:

$\cos^2\Big(\frac\pi6+\text{x}\Big)-\sin^2\Big(\frac{\pi}{6}-\text{x}\Big)$

$=\cos\Big(\frac{\pi}{6}+\text{x}+\frac{\pi}{6})-\text{x}\Big)\cos\Big(\frac{\pi}{6}+\text{x}-\frac{\pi}{6}+\text{x}\Big)$  $\Big[\text{Using}\cos(\text{A+B})\cos(\text{A-B})=\cos^2\text{A}-\sin^2\text{B}\Big]$

$=\cos\frac{2\pi}{6}\cos2\text{x}$

$=\frac12\cos2\text{x}$ $\Big[\text{As}\cos\frac\pi3=\frac12\Big]$

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. $-2\sec\text{x}$

Solution:

$-2\sec\text{x}$

$\sqrt{\frac{1-\sin\text{x}}{1+\sin\text{x}}}+\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}$

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

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

$=\sqrt{\frac{(1-\sin\text{x})^2}{\cos^2\text{x}}}+\sqrt{\frac{(1+\sin\text{x})^2}{\cos^2\text{x}}}$

$=\frac{(1-\sin\text{x})}{-\cos\text{x}}+\frac{(1+\sin\text{x})}{-\cos\text{x}}$ $[\frac{\pi}{2}<\text{x}<\pi,\text{so}\cos\text{x}\text{ will }\text{be }\text{negative}.]$

$-(\sec\text{x}-\tan\text{x})-(\sec\text{x}+\tan\text{x})$

$=-2\sec\text{x}$

4. 0

Solution:

Given expression is $(\sin45^\circ+\theta)-\cos(45^\circ-\theta)$

$\sin(45^\circ+\theta)=\sin45^\circ\cos\theta+\cos45^\circ\sin\theta$

$=\frac{1}{\sqrt2}\cos\theta+\frac{1}{\sqrt2}\sin\theta$

$\cos(45^\circ-\theta)=\cos45^\circ\cos\theta+\sin45^\circ\sin\theta$

$$$=\frac{1}{\sqrt2}\cos\theta+\frac{1}{\sqrt2}\sin\theta$

$\sin(45^\circ+\theta)-\cos(45^\circ-\theta)$

$=\frac{1}{\sqrt2}\cos\theta+\frac{1}{\sqrt2}\sin\theta-\frac{1}{\sqrt2}\cos\theta-\frac{1}{\sqrt2}\sin\theta$

= 0. Hence, the correct option is (d). 

4. $-1-\cot\text{a}$

Solution:

We have:

$\sqrt{2\cot\alpha+\frac{1}{\sin\alpha}}$

$=\sqrt{\frac{2\cos\alpha}{\sin\alpha}+\frac{1}{\sin^2\alpha}}$

$=\sqrt{\frac{2\sin\alpha\cos\alpha+1}{\sin\alpha}}$

$=\sqrt{\frac{2\sin\alpha\cos\alpha+\sin^2\alpha+\cos^2\alpha}{\sin\alpha}}$

$=\sqrt{\frac{(\sin\alpha+\cos\alpha)^2}{\sin^2\alpha}}$

$=\sqrt{(1+\cot\alpha)^2}$

$=|1+\cot\alpha|$

$=-(1+\cot\alpha) $ $$ $[\text{ when}\frac{3\pi}{4}<\alpha<\pi,\cot\alpha<-1\Rightarrow\cot\alpha+1<0\big]$

$=-1-\cot\alpha$

2. $\sin4\beta$

Solution:

it is given thet $\tan\alpha=\frac{1}{7}$ and $\tan\beta=\frac{1}{3}.$

Now,

$\tan\beta=\frac{2\tan\beta}{1-\tan^2\beta}$

$=\frac{2\times\frac{1}{3}}{1-\frac{1}{9}}$

$=\frac{\frac{2}{3}}{\frac{8}{9}}$

$=\frac{3}{4}$

$\therefore\tan(\alpha+2\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$

$=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7}\times\frac{3}{4}}$

$=\frac{\frac{25}{28}}{\frac{25}{28}}$

$=1$

$\tan(\alpha+2\beta)=1=\tan\frac{\pi}{4}$

$\Rightarrow\alpha+2\beta=\frac{\pi}{4}$

$\Rightarrow\alpha=\frac{\pi}{4}-2\beta$

$\Rightarrow2\alpha=\frac{\pi}{2}-4\beta$

$\Rightarrow\cos2\alpha=\cos\big(\frac{\pi}{2}-4\beta\big)=\sin4\beta$

$\therefore\cos2\alpha=\sin4\beta$

Hence, the correct answer is option B.

4.  $\frac{\cos\text{x}-\text{e}}{1-\text{e}\cos\text{x}}$

Solution:

Given:

$\tan\frac{​​\text{x}}{2}=\sqrt{\frac{1-\text{e}}{1+\text{e}}}\tan\frac{\alpha}{2}$

$\Rightarrow\frac{\tan\frac{\text{x}}{2}}{\tan\frac{\alpha}{2}}=\sqrt{\frac{1-\text{e}}{1+\text{e}}}$

Squaring both sides, we get,

$\frac{\tan^2\frac{\text{x}}{2}}{\tan^2\frac{\alpha}{2}}=\frac{1-\text{e}}{1+\text{e}}$

$\Rightarrow\tan^2\frac{\alpha}{2}(1-\text{e})=\tan^2\frac{\text{x}}{2}(1+\text{e})$

$\Rightarrow\frac{\sin^2\frac{\alpha}{2}}{\cos^2\frac{\alpha}{2}}(1-\text{e})=\frac{\sin^2\frac{\text{x}}{2}}{\cos^2\frac{\text{x}}{2}}(1+\text{e})$

$\Rightarrow\frac{\frac{1}{2}(1-\cos\alpha)}{\frac{1}{2}(1+\cos\alpha)}(1-\text{e})=\frac{\frac{1}{2}(1-\cos\text{x})}{\frac{1}{2}(1+\cos\text{x})}(1+\text{e})$

$\Rightarrow(1-\cos\alpha)(1+\cos\text{x})(1-\text{e})=(1+\cos\alpha)(1-\cos\text{x})(1+\text{x})$

$\Rightarrow(1+\cos\text{x})(1-\text{e})-\cos\alpha(1+\cos\text{x})(1-\text{e})\\=(1-\cos\text{x})(1+\text{e})+\cos\alpha(1-\cos\text{x})(1+\text{e})$

$\Rightarrow\cos\alpha\big\{(1+\cos\text{x})(1-\text{e})+(1-\cos\text{x})(1+\text{e})\big\}\\=(1+\cos\text{x})(1-\text{e})-(1-\cos\text{x})(1+\text{e})$

$\Rightarrow\cos\alpha=\frac{2\cos\text{x}-2\text{e}}{2-2\text{e}\cos\text{x}}=\frac{\cos\text{x}-\text{e}}{1-\text{e}\cos\text{x}}$ 

2. $\frac{\pi}{3}$

Solution:

Let the angles of the triangle be (a - d), (a) and (a + d).

Thus, we have:

a - d + a + a + d = 180

⇒ 3a = 180

⇒ a = 60

2. 2

Solution:

We have,

$\sin^2\Big(\frac{\pi}{18}\Big)+\sin^2\Big(\frac{7\pi}{18}\Big)+\sin^2\Big(\frac{4\pi}{9}\Big)$

$=\frac{1}{2}\Big[1-\cos\big(\frac{\pi}{9}\big)+1-\cos\big(\frac{2\pi}{9}\big)+1-\cos\frac{7\pi}{9}+1-\cos\frac{8\pi}{9}\Big]$ $\Big(\therefore\sin^2\theta=\frac{1-\cos2\theta}{2}\Big)$

$=\frac{1}{2}\Big[4-\cos\big(\frac{\pi}{9}\big)-\cos\big(\frac{2\pi}{9}\big)-\Big\{-\cos\Big(\pi-\frac{7\pi}{9}\Big)\Big\}-\big\{-\cos(\pi-\frac{8\pi}{9})\big\}\Big]$

$=\frac{1}{2}\Big[4-\cos\big(\frac{\pi}{9}\big)-\cos\big(\frac{2\pi}{9}\big)+\cos\big(\frac{2\pi}{9}\big)+\cos\big(\frac{\pi}{9}\big)\Big]$

$=\frac{4}{2}$

$=2$

2. $\frac{2}{\text{e}^\text{x}+\text{e}^{-\text{x}}}$

Solution:

We have:

$\tan\theta + \sec\theta=\text{e}^\text{x}$

$\sec\theta + \tan\theta = \text{e}^\text{x}\cdots(1)$

$\Rightarrow\frac{1}{\sec\theta+\tan\theta}=\frac{1}{\text{e}^\text{x}}$

$\Rightarrow\frac{\sec^2\theta-\tan^2\theta}{\sec\theta + \tan\theta}=\frac{1}{\text{e}^\text{x}}$

$\Rightarrow\frac{(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{(\sec\theta + \tan\theta)}=\frac{1}{\text{e}^\text{x}}$

$\therefore \sec\theta - \tan\theta=\frac{1}{\text{e}^\text{x}}\cdots(2)$

Adding (1) and (2): 

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

$\Rightarrow 2\sec\theta = \frac{(\text{e}^\text{x})^2+1}{\text{e}^\text{x}}$

$\Rightarrow \sec\theta = \frac{\text{e}^{2\text{x}}+1}{2\text{e}^\text{x}}$

$\Rightarrow\sec\theta=\frac{1}{2}\times\frac{\text{e}^{2\text{x}}+1}{2\text{e}^\text{x}}$

$\Rightarrow\sec\theta = \frac{1}{2}\times(\text{e}^\text{x}+\text{e}^\text{-x})$

$\Rightarrow\frac{1}{\cos\theta}=\frac{\text{e}^\text{x}+\text{e}^\text{x}}{2}$

$\Rightarrow\cos\theta= \frac{2}{\text{e}^\text{x}+\text{e}^\text{-x}}$

3. $\frac{\pi}{6}$

Solution:

We have:

$\tan\text{x}+\sec\text{x}=\sqrt{3}$ $[0,<\text{x}<\pi]$

$\Rightarrow\sec\text{x}+\tan\text{x}=\sqrt{3}$

$\Rightarrow\frac{1}{\cos\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}=\sqrt{3}$

$\Rightarrow1+\sin\text{x}=\sqrt{3}\cos\text{x}$

$\Rightarrow(1+\sin\text{x})^2=(\sqrt{3}\cos\text{x})^2$

$\Rightarrow1+\sin^2\text{x}+2\sin\text{x}=3\cos^2\text{x}$

$\Rightarrow1+\sin^2\text{x}+2\sin\text{x}=3(1-\sin^2\text{x})$

$\Rightarrow4\sin^2\text{x}+2\sin\text{x}=2$

$\Rightarrow2\sin^2\text{x}+\sin\text{x}-1=0$

$\Rightarrow\sin\text{x}=-1,\frac{1}{2}$

since $0<\text{x}<\pi,\ \sin\text{x}$ cannot be negative.

$\therefore\sin\text{x}=\frac{1}{2}$

$\therefore\text{x}=\frac{\pi}{6}$

3. 1

Solution:

$\cot\Big(\frac{\pi}{4}+\theta\Big)\cdot\cot\Big(\frac{\pi}{4}-\theta\Big)=\frac{\cot\frac{\pi}{4}\cot\theta-1}{\cot\theta+\cot\frac{\pi}{4}}\times\frac{\cot\frac{\pi}{4}\cot\theta+1}{\cot\theta-\cot\frac{\pi}{4}}$

$=\frac{1.\cot\theta-1}{\cot\theta+1}\times\frac{1.\cot\theta+1}{\cot-1}$

$=\frac{\cot\theta-1}{\cot\theta+1}\times\frac{\cot\theta+1}{\cot\theta-1}=1$

Hence, the correct option is (c).

3. $\cos2\text{A}$

Solution:

We have,

$2\sin^2​\text{B}+4\cos(\text{A+B})\sin​​\text{A}\sin​\text{B}+\cos2​\text{A+B}$

$=1-\cos2​\text{B}+\cos^2(​\text{A+B})+4\cos(​\text{A+B})\sin​\text{A}\sin​\text{B}$

$=1+(\cos2(​\text{A+B})-\cos2​\text{B})+4\cos(​\text{A+B})\sin​\text{A}\sin​\text{B}$

$=1-2\sin​\text{A}\sin(​\text{A+2B})+4\cos(​\text{A+B})\sin​\text{A}\sin​\text{B}\\ \ \ \ \ \ \ \ \ \Big[\because\cos​\text{C}\cos​\text{D}=-2\sin\frac{​\text{C+D}}{2}\sin\frac{​\text{C}-​\text{D}}{2}\Big]$

$=1-2\sin​\text{A}[\sin(​\text{A}+2​\text{B})-2\sin​\text{B}-2\cos(​\text{A+B})]$

$=1-2\sin​\text{A}[\sin(​\text{A}+2​\text{B})-\big\{\sin(\text{B+A+B})+\sin(\text{B}-\text{(A+B)})\big\}\\ \ \ \ \ \ [\because2\sin\text{C}\cos\text{D}=\sin(\text{C+D})+\sin(\text{C}-\text{D})]$

$=1-2\sin\text{A}\big[\sin(\text{A+2B})-\big\{\sin(\text{A+2B})+\sin(-\text{A})\big\}\big]$

$=1-2\sin\text{A}[\sin\text{A}]$

$=1-2\sin^2\text{A}$

$=\cos^2\text{A}$

1. $\tan\text{A}\tan\text{B}<1$

1. $\frac{\tan\alpha}{2}+\frac{\tan\beta}{2}+\frac{\tan\text{y}}{2}=\frac{\tan\alpha}{2}.\frac{\tan\beta}{2}.\frac{\tan{\text{y}}}{2}$

1. $\frac{\pi}{3}$

Solution:

Given: $\cos\text{x}+\sqrt{3}\sin\text{x}=2\ ......(1)$

This equation is of the form $\text{a}\cos\text{x}+\text{b}\sin\text{x}=\text{c},$ where $\text{a}=1,\ \text{b}=\sqrt{3}$ and $\text{c}=2.$

Let:

$\text{a}=\text{r}\ \cos\alpha$ and $\text{b}=\sin\alpha$

Now,

$1=\text{r}\ \cos\alpha,\sqrt{3}=\text{r}\sin\alpha$

$\Rightarrow\text{r}=\sqrt{\text{a}^2+\text{b}^2}=\sqrt{1+3}=\sqrt{4}=2$

And,

$\tan\alpha=\frac{\text{b}}{\text{a}}$

$\Rightarrow\tan\alpha=\frac{\sqrt{3}}{1}$

$\Rightarrow\tan\alpha\sqrt{3}$

$\Rightarrow\alpha=\frac{\pi}{3}$

On putting $\text{a}=1=\text{r}\ \cos\alpha$ and $\text{b}=\sqrt{3}=r\ \sin\alpha$ in equation (1), we get:

$\Rightarrow\text{r}\cos(\text{x}-\alpha)=2$

$\Rightarrow2\cos(\text{x}-\frac{\pi}{3})=2$

$\Rightarrow\cos(\text{x}-\frac{\pi}{3})=1$

$\Rightarrow\cos(\text{x}-\frac{\pi}{3})=\cos0$

$\Rightarrow\text{x}=\frac{\pi}{3}=2\text{n}\pi\pm0$

$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3}$

For $\text{n}=0,\text{x}=\frac{\pi}{3}.$

$\therefore\text{x}=\frac{\pi}{3}$

2. $\text{n}\pi+(-1)^\text{n}\frac{\pi}{6},\text{n}\in\text{Z}$

Solution:

Given equation:

$\cot\text{x}-\tan\text{x}=\sec\text{x}$

$\Rightarrow\frac{\cos\text{x}}{\sin\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}=\frac{1}{\cos\text{x}}$

$\Rightarrow\frac{\cos^2\text{x}-\sin^2}{\sin\text{x}\cos​x​}=\frac{1}{\cos\text{x}}$

$\Rightarrow\cos^2​\text{x}​\sin^2=\sin\text{x}$

$\Rightarrow(1-\sin^2\text{x})-\sin^2\text{x}=\sin\text{x}$

$\Rightarrow1-2\sin^2=\sin\text{x}$

$\Rightarrow2\sin^2\text{x}+\sin\text{x}-1=0$

$\Rightarrow2\sin^2\text{x}+2\sin\text{x}-\sin\text{x}-1=0$

$\Rightarrow2\sin\text{x}(\sin\text{x}+1)-1(\sin\text{x}+1)=0$

$\Rightarrow(\sin\text{x}+1)(2\sin\text{x}-1)=0$

$\Rightarrow\sin\text{x}+1=0$ or $2\sin\text{x}-1=0$

$\Rightarrow\sin\text{x}=-1$ or $\sin\text{x}=\frac{1}{2}$

Now,

$\sin\text{x}=\frac{1}{2}\Rightarrow\sin\text{x}=sin\frac{3\pi}{2}\Rightarrow\text{x}=\text{m}\pi+(-1)^\text{m}\frac{3\pi}{2},\ \text{m}\in\text{Z}$

And

$\sin\text{x}=\frac{1}{2}\Rightarrow\sin\text{x}=\sin\frac{\pi}{6}\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6},\text{n}\in\text{Z}$

$\therefore\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6},\text{n}\in\text{Z}$

2. $\tan2\alpha=\tan\beta$

Solution:

$\tan\alpha=\frac{1-\cos\beta}{\sin\beta}$

$=\frac{2\sin^2\frac{\beta}{2}}{2\sin\frac{\beta}{2}\cos\frac{\beta}{2}}$

$=\frac{\sin\frac{\beta}{2}}{\cos\frac{\beta}{2}}$

$\Rightarrow\tan\alpha=\tan\frac{\beta}{2}$

$\Rightarrow\alpha=\frac{\beta}{2}$

$\Rightarrow2\alpha=\beta$

$\therefore\tan2\alpha=\tan\beta$

3. $\frac{\sqrt{3}}{2}$

Solution:

Given that, $\frac{1-\tan^215^\circ}{1+\tan^215^\circ}$

Let $\theta=15^\circ\therefore2\theta=30^\circ$

$\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$

$\Rightarrow\cos30^\circ=\frac{1-\tan^215^\circ}{1+\tan^215^\circ}\Rightarrow\frac{\sqrt3}{2}=\frac{1-\tan^215^\circ}{1+\tan^215^\circ}$

Hence, the correct option is (c).

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

Solution:$$

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

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

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

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

$=\ \sin30^\circ$

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

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

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

$=\ 0$