MATHS M.C.Q (1 Marks)

4. $\Big(\frac{5{\pi}}{4},\frac{7\pi}{4}\Big)$

2. $0$

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

Solution:

Given:

$\sqrt{3}(\cot\text{x}+\tan\text{x})=4$

$\Rightarrow\sqrt{3}\Big(\frac{\cos\text{x}}{\sin\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}\Big)=4$

$\Rightarrow\sqrt{3}(\cos^2\text{x}+\sin^2\text{x})=4\sin\text{x}\cos\text{x}$

$\Rightarrow\sqrt{3}=2\sin2\text{x}$ $\big[\sin2\text{x}=2\sin\text{x}\cos\text{x}\big]$

$\Rightarrow\sin2\text{x}=\frac{\sqrt{3}}{2}$

$\Rightarrow\sin2\text{x}=\sin\frac{\pi}{3}$

$\Rightarrow2\text{x}=\text{n}\pi+(-1)^{\text{n}}\frac{\pi}{3},\text{n}\in\text{Z}$

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

To obtain the smallest value of x, we will put n = 0n = 0 in the above equation.

Thus, we have:

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

Hence, the smallest value of x is $\frac{\pi}{6}.$

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

Solution:

$\tan135^\circ=\tan(90^\circ+45^\circ)$

$=-\tan45^\circ$

$=-1$

Or, $\tan(69^\circ+66^\circ)=\frac{\tan69^\circ+\tan66^\circ}{1-\tan69^\circ\tan66^\circ}$

$\Rightarrow-1=\frac{\tan69^\circ+\tan66^\circ}{1-\tan69^\circ\tan66^\circ}$

$\Rightarrow\tan69^\circ+\tan66^\circ-\tan69^\circ+\tan66^\circ=-1$

$\therefore2\text{k}=-1$

$\Rightarrow\text{k}=\frac{-1}{2}$

3. 9.5

Solution:

We have:

$\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+...+\sin^285^\circ+\sin^290^\circ$

$=\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+...+\sin^2(90^\circ-10^\circ)+\sin^2(90^\circ-5^\circ)+\sin^290^\circ$

$=\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+...+\cos^210^\circ+\cos^25^\circ+\sin^290^\circ$

$=(\sin^25^\circ+\cos^25^\circ)+(\sin^210^\circ+\cos^210^\circ)+(\sin^215^\circ+\cos^215^\circ)$

$+(\sin^220^\circ+\cos^220^\circ)+(\sin^225^\circ+\cos^225^\circ)+(\sin^230^\circ+\cos^230^\circ)$

$+(\sin^235^\circ+\cos^235^\circ)+(\sin^240^\circ+\cos^240^\circ)+\sin^245^\circ+\sin^290^\circ$

$=1+1+1+1+1+1+1+1+\Big(\frac{1}{\sqrt2}\Big)+(1)^2$ $[\because \sin^2\theta+\cos^2\theta=1]$

$=8+\frac{1}{2}+1$

$=9.5$

1. $0$

Solution:

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

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

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

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

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

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

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

$=\ 0$

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

Solution:

$\cos12^\circ+\cos84^\circ+\cos156^\circ+\cos132^\circ$

$=(\cos12^\circ+\cos132^\circ)+(\cos84^\circ+\cos156^\circ)$

$=2\cos72^\circ\cos60^\circ+2\cos120^\circ\cos36^\circ$

$=\cos72^\circ-\cos36^\circ=\sin18^\circ-\cos36^\circ$

$=\Big(\frac{\sqrt5-1}{4}\Big)-\Big(\frac{\sqrt5+1}{4}\Big)=\frac{-1}{2}$

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

3. 2

Solution:

$\sin\theta+\text{cosec}\theta=2$

$\Rightarrow\sin\theta+\frac{1}{\sin\theta}=2$

$\Rightarrow\sin^2\theta+1=2\sin\theta\Rightarrow(\sin\theta-1)^2=0\Rightarrow\sin\theta=1$

$\therefore\sin^2\theta+\text{cosec}^2\theta=1+1=2$

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

Solution:

We have:

$\sec\text{x} = \text{x} +\frac{1}{4\text{x}}$

$\Rightarrow\sec^2\text{x}=\text{x}^2+\frac{1}{16\text{x}^2}+\frac{1}{2}$

$\Rightarrow1+\tan^2\text{x}$

$=1+\text{x}^2+\frac{1}{16\text{x}^2}-\frac{1}{2}$

$\Rightarrow\tan^2\text{x}=\text{x}^2+\frac{1}{16\text{x}^2}-\frac{1}{2}$

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

$\therefore\tan\text{x}=\pm\Big(\text{x}-\frac{1}{4\text{x}}\Big)$

$\sec\text{x}-\tan\text{x}=\Big(\text{x}+\frac{1}{4\text{x}}\Big)-\Big(\text{x}-\frac{1}{4\text{x}}\Big)\text{or}$ 

$=\Big(\text{x}+\frac{1}{4\text{x}}\Big)-\Big[-\Big(\text{x}-\frac{1}{4\text{x}}\Big)\Big]$

$=\frac{1}{2\text{x}}\text{ or } 2\text{x}$

1. $\frac{\sqrt{5}+1}{8}$

Solution:

Given expression is $\cos^248^\circ-\sin^212^\circ$

$\cos^248^\circ-\sin^212^\circ=\cos(48^\circ+12^\circ).\cos(48^\circ-12^\circ)$

$[\therefore\cos^2\text{A}-\sin^2\text{B}=\cos(\text{A+B}).\cos(\text{A}-\text{B})]$

$=\cos60^\circ.\cos36^\circ=\frac{1}{2}\times\frac{\sqrt5+1}{4}=\frac{\sqrt5+1}{8}$

Hence,the correct option is (a).

2. $\tan\text{x}$

Solution:

We have,

$\cot​​\text{x}-2\cot2\text{x}=\cot\text{x}-2\frac{\cot^2\text{x}-1}{2\cot\text{x}}$

$=\frac{1}{\cot\text{x}}$

$=\tan\text{x}$

4. $\frac\pi4$

Solution:.

$\tan(\text{A+B})=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}$

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

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

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

$=1$

$\therefore \text{ A+B}=\tan^{-1}(1)=\frac\pi4.$

1. $-2\text{x},\frac{1}{2\text{x}}$

Solution:

We have:

$\tan\text{x}=\text{x}-\frac{1}{4\text{x}}$

$\Rightarrow\sec^2\text{x}=1+\tan^2\text{x}$

$\Rightarrow\sec^2\text{x}=1+\Big(\text{x}-\frac{1}{4\text{x}}\Big)^2$

$\Rightarrow\sec^2\text{x}=\text{x}^2+\frac{1}{16\text{x}^2}+\frac{1}{2}$

$\Rightarrow \sec\text{x}^2 =\Big(\text{x}+\frac{1}{4\text{x}}\Big)^2$

$\therefore \sec\text{x}=\pm\Big(\text{x}+\frac{1}{4\text{x}}\Big)$

$\Rightarrow \sec\text{x}-\tan\text{x} =\Big(\text{x}+\frac{1}{4\text{x}}\Big)- \Big(\text{x}-\frac{1}{4\text{x}}\Big)\text{ or} -\Big(\text{x}+\frac{1}{4\text{x}}\Big)\text{or}- \Big(\text{x}-\frac{1}{4\text{x}}\Big)$

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

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

Solution:

 Given that, $\tan\alpha=\frac{\text{m}}{\text{m}+1}$ and $\tan\beta=\frac{1}{2\text{m}+1}$

Now, $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\cdot\tan\beta}=\frac{\frac{\text{m}}{\text{m}+1}+\frac{1}{2\text{m}+1}}{1-\Big(\frac{\text{m}}{\text{m}+1}\Big)\Big(\frac{1}{2\text{m}+1}\Big)}$

$=\frac{\text{m}(2\text{m}+1)+\text{m}+1}{(\text{m}+1)(2\text{m}+1)-\text{m}}=\frac{2\text{m}^2+2\text{m}+1}{2\text{m}^2+3\text{m}+1-\text{m}}=1$

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

2. 2

Solution:

Given that, $\alpha+\beta=\frac{\pi}{4}\Rightarrow\tan(\alpha+\beta)=\tan\frac{\pi}{4}$

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

$\Rightarrow\tan\alpha+\tan\beta=1-\tan\alpha\tan\beta$

$\Rightarrow\tan\alpha+\tan\beta+\tan\alpha\tan\beta=1$

On adding 1 both sides, we get,

$\Rightarrow1+\tan\alpha+\tan\beta+\tan\alpha\tan\beta=1+1$ 

$\Rightarrow1(1+\tan\alpha)+\tan\beta(1+\tan\alpha)=2$

$\Rightarrow(1+\tan\alpha)(1+\tan\beta)=2$

Hence, the correct option is (b)

2. $\text{x}=\frac{2\pi}{3},\ \frac{4\pi}{3}$

Solution:

Given equation:

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

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

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

or

$\cos\text{x}=\cos\frac{4\pi}{3}$

$\Rightarrow\text{x}=\frac{4\pi}{3}$

so, both $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ line in $0<\text{}\text{x}<2\pi.$

2. $\sqrt{6}+\frac{\sqrt{2}}{4}$

2. $1$

Solution:

Given,

$(2^\text{n}+1)\text{x}=\pi$

$\Rightarrow2^\text{n}\text{x}+\text{x}=\pi$

$\Rightarrow2^\text{n}\text{x}=\pi-\text{x}$

$\Rightarrow\sin2^\text{n}\text{x}=\sin(\pi-\text{x})$

$\Rightarrow\sin2^\text{n}\text{x}=\sin\text{x}\ .....(1)$

$2^\text{n}\cos\text{x}\cos2\text{x}\cos2^2\text{x}\ ...\cos2^{\text{n}-1}\text{x}=2^\text{n}\times\frac{\sin2^\text{n}\text{x}}{2^\text{n}\sin\text{x}}$

$=\frac{\sin2^\text{n}\text{x}}{\sin\text{x}}$

$=\frac{\sin\text{x}}{\sin\text{x}}$ [From (1)]

$=1$

4.  $\frac{\pi}{3}$

Solution:

Given equation:

$\cot\text{x}+\sqrt{3}\sin​​​​\text{x}=2\ .....(1)$

Thus, the 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 $\text{c}=3$

Let:

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

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

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

$\tan\alpha=\frac{\text{b}}{\text{a}}\Rightarrow\tan\alpha=\frac{\sqrt{3}}{1}\Rightarrow\tan\alpha\tan\frac{\pi}{3}\Rightarrow\alpha=\frac{\pi}{3}$

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

$\text{r}\cos\alpha\cos\text{x}+\text{r}\sin\alpha\sin\text{x}=2$

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

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

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

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

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

$\Rightarrow​\text{x}​-\frac{\pi}{3}=0$

$\Rightarrow\text{x}=\frac{\pi}{3}$ 

3. $\tan\text{x}-\sec\text{x}$

Solution:

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

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

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

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

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

$=\frac{(1-\sin\text{x})}{-\cos\text{x}}$ $[\text{as},\frac{\pi}{2}<\text{x}<\frac{3\pi}{2},\text{so}\cos\theta \text{ will}\text{ be}\text{ negative}]$

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

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

2. 0

Solution:

since $\cos90^\circ=0,$ we have

$\cos1^\circ\cos2^\circ\cos3^\circ\dots\cos90^\circ\dots\cos179^\circ=0$

4. $\text{n}\pi+(-1)^\text{n}\frac{\pi}{4}-\frac{\pi}{3},\ \text{n}\in\text{Z}$

Solution:

Given equation:

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

Thus, is of the  form $\text{a}\cos\text{x}+\text{b}\sin\text{x}=\text{c},$ where $\text{a}=\sqrt{3},\ \text{b}=1$ and $\text{c}=\sqrt2$

Let:

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

Now,

$\text{r}=\sqrt{\text{a}^2+\text{b}^2}=\sqrt{\big(\sqrt{3}\big)^2+1^2}=2$ 

And

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

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

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

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

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

$\text{r}\cos\text{x}\sin\alpha+\text{r}\sin\text{x}\cos\alpha=\sqrt{2}$

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

$\Rightarrow2\sin(\text{x}+\alpha)=\sqrt{2}$

$\Rightarrow\sin\big(\text{x}+\frac{\pi}{3}\big)=\frac{1}{\sqrt{2}}$

$\Rightarrow\sin\big(\text{x}+\frac{\pi}{3}\big)=\cos\frac{\pi}{4}$

$\Rightarrow\text{x}\frac{\pi}{3}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{4},\ \text{n}\in\text{Z}$

$\Rightarrow\text{x}​​\text{n}\pi+(-1)^\text{n}\frac{\pi}{4}-\frac{\pi}{3},\ \ \text{n}\in\text{Z}$

1. $\tan3\text{x}$

Solution:

$\frac{\pi}{3}=60^\circ$

$\tan​​\text{x}\tan(60^\circ-\text{x})\tan(60^\circ+\text{x})\\=\tan\text{x}\times\frac{\tan60^\circ-\tan\text{x}}{1+\tan60^\circ\tan\text{x}}\times\frac{\tan60^\circ+\tan\text{x}}{1-\tan60^\circ\tan\text{x}}$

$=\tan\text{x}\times\frac{\sqrt{3}-\tan\text{x}}{1+\sqrt{3\tan\text{x}}}\times\frac{\sqrt{3}+\tan\text{x}}{1-\sqrt{3}\tan\text{x}}$

$=\frac{\tan\text{x}(3-\tan^2\text{x})}{1-3\tan^2\text{x}}$

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

$=\tan3\text{x}$

1. $\cos2\text{A}$

Solurion:

$\cos(36^\circ-​\text{A})\cos(36^\circ+\text{A})+\cos(54^\circ-\text{A})\cos(54^\circ+\text{A})$

$=\cos[90^\circ-(54^\circ+\text{A})]\cos[90^\circ-(54^\circ-\text{A})]+\cos(54^\circ\\-\text{A})\cos(54^\circ+\text{A})$

$=\sin(54^\circ+\text{A})\sin(54^\circ-\text{A})+\cos(54^\circ-\text{A})\cos(54^\circ+\text{A})$ $[\cos(90^\circ-\theta)=\sin\theta]$

$=\cos(54^\circ+\text{A}-54^\circ+\text{A})$ $[\cos(\text{A}-\text{B})=\cos\text{A}\cos\text{B}+\sin\text{A}\sin\text{B}]$

$=\cos2\text{A}$

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

Solution:

Given:

$\sin\alpha+\sin\beta=\text{a}$

$\Rightarrow2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha+\beta}{2}=\text{a}\ .....(1)$

Also,

$\cos\alpha+\cos\beta=\text{b}$

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

On dividing (1) by (2), we get

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

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

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

1. 2

Solution:

$\tan(\text{A - B})=\tan\frac\pi4$

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

$\Rightarrow\tan\text{A}-\tan\text{B}=1+\tan\text{A}\tan\text{B}\cdots(1)$

Now,

$(1+\tan\text{A})(1-\tan\text{B})=1+\tan\text{A}-\tan\text{B}-\tan\text{A}\tan\text{B}$

$=1+1+\tan\text{A}\tan\text{B}-\tan\text{A}\tan\text{B}$

$=2$

1. $\tan3\text{A}\tan2\text{A}\tan\text{A}$

Solution:

$3\text{A}=\text{A}+2\text{A}$

$\Rightarrow\tan3\text{A}=\tan(\text{A+2A})$

$\Rightarrow\tan3\text{A}=\tan\text{A}+\tan\frac{2\text{A}}{1}-\tan\text{A}.\tan2\text{A}$

$\Rightarrow\tan\text{A}+\tan2\text{A}=\tan3\text{A}-\tan3\text{A}.\tan2\text{A}.\tan\text{A}$

$\Rightarrow\tan3\text{A}-\tan2\text{A}-\tan\text{A}=\tan3\text{A}.\tan2\text{A}.\tan\text{A}$

3. $\sec\text{x}=\frac12$

Solution:

1.  $\sin\text{x} = -\frac{1}{5}$ is correct as $-1\leq \sin\text{x} \leq1$

2. $\cos\text{x}=1$ is correct as $-1\leq \cos\text{x}\leq1$

3. $\sec\text{x}=\frac{1}{2}$ is correct as $\text{x}\in [(-\infty, -1)\ \cup (1,\infty)]$

4. $\tan\text{x} = 20$ is correct as $\tan\text{x}$ can take any real value.

Hence, the correct answer is option C.

1. $\theta,\phi$

Solution:

We have:

$\text{x}=\text{r}\sin\theta\cos\phi,\text{y}=\text{r}\sin\theta\sin\phi\text{ and }\text{z}=\text{r}\cos\theta,$

$\therefore\text{x}^2+\text{y}^2+\text{z}^2$

$=(\text{r}\sin\theta\cos\phi)^2+(\text{r}\sin\theta\sin\phi)^2+(\text{r}\cos\theta)^2$

$=\text{r}^2\sin^2\theta\cos^2\phi+\text{r}^2\sin^2\theta\sin^2\phi+\text{r}^2\cos^2\theta$

$=\text{r}^2\sin^2\theta(\cos^2\phi+\sin^2\phi)+\text{r}^2\cos^2\theta$

$=\text{r}^2\sin^2\theta\times1+\text{r}^2\cos^2\theta$

$=\text{r}^2\sin^2\theta+\text{r}\cos^2\theta$

$=\text{r}^2(\sin^2\theta+\cos^2\theta)$

$=\text{r}^2\times1$

$=\text{r}^2$

Thus, $\text{x}^2+\text{y}^2+\text{z}^2$ is independent of $\theta\text{ and }\phi$

3. $\text{n}\pi\pm\frac{\pi}{6},\text{n}\in\text{Z}$

Solution:

Given:

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

$\Rightarrow\sin^2\text{x}=\frac{1}{4}$

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

$\Rightarrow\sin\text{x}=\sin\frac{\pi}{6}$ or $\sin\text{x}=\sin\Big(-\frac{\pi}{6}\Big)$

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

$\Rightarrow\text{x}=\text{n}\pi\pm\frac{\pi}{6},\text{n}\in\text{Z}$

1. $2\sqrt{3}$

Solution:

$\tan75^\circ-\cot75^\circ=\frac{\sin75^\circ}{\cos75^\circ}-\frac{\cos75^\circ}{\sin75^\circ}\\=\frac{2(\sin^275^\circ-\cos^275^\circ)}{2\sin75^\circ\cos75^\circ}=\frac{-2\cos150^\circ}{\sin150^\circ}$ 

$=-2\cot150^\circ=-2\cot(180^\circ-30^\circ)=2\cot30^\circ=2\sqrt3$

3. $\text{cosec}\ \text{x}$

Solution:

We have,

$\frac{2(\sin2​\text{x}+2\cos^2\text{x}-1​)}{\cos\text{x}-\sin\text{x}-\cos3\text{x}+\sin3\text{x}}$

$=\frac{2(\sin2\text{x}+\cos2\text{x})}{\cos\text{x}-\sin\text{x}-4\cos^3\text{x}+3\cos\text{x}\sin\text{x}-4\sin^3\text{x}}$

$=\frac{2(\sin2\text{x}+\cos2\text{x})}{4\cos​​\text{x}-4\cos^3\text{x}+2\sin\text{x}-4\sin^3\text{x}}$

$=\frac{2(\sin2\text{x}+\cos2\text{x})}{4\cos​​\text{x}(1-\cos^2​​​​\text{x})+2\sin\text{x}(1-2\sin^2\text{x})}$

$=\frac{2(\sin2\text{x}+\cos2\text{x})}{4\cos​​\text{x}\sin^2\text{x}+2\sin\text{x}\cos2\text{x}}$

$=\frac{2(\sin2\text{x}+\cos2\text{x})}{2\times2\sin\text{x}\cos\text{x}\sin\text{x}+2\sin\text{x}\cos2\text{x}}$

$=\frac{2(\sin2\text{x}+\cos2\text{x})}{2\sin2\text{x}\sin2\text{x}+2\sin\text{x}\cos2\text{x}}$

$=\frac{2(\sin2\text{x}+\cos2\text{x})}{2\sin\text{x}(\sin2\text{x}+\cos2\text{x})}$

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

$=\text{cosec}\ \text{x}$

2. 2

Solution:

$\sin^2\text{x}-\cos\text{x}=\frac{1}{4}$

$\Rightarrow(1-\cos^2\text{x})-\cos\text{x}=\frac{1}{4}$

$\Rightarrow4-4\cos^2\text{x}-4\cos\text{x}=1$

$\Rightarrow4\cos^2-4\cos\text{x}=1$

$\Rightarrow4\cos^2\text{x}+4\cos\text{x}-3=0$

$\Rightarrow4\cos^2\text{x}+6\cos\text{x}-2\cos\text{x}-3=0$

$\Rightarrow2\cos\text{x}(2\cos\text{x}+3)-1(2\cos\text{x}-1)=0$

$\Rightarrow(2\cos\text{x}+3)(2\cos\text{x}-1)=0$

$\Rightarrow2\cos\text{x}+3=0$ or $2\cos\text{x}-1=0$

$\Rightarrow\cos\text{x}=-\frac{3}{2}$  is not possible.

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

$\Rightarrow\cos\text{x}=\cos\frac{\pi}{3}$

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

Now for n = 0 and 1, the values of x are $\frac{\pi}{3},\ \frac{5\pi}{3}$ and $\frac{7\pi}{3},$ but $\frac{7\pi}{3}$  is not in $[0,\ 2\pi]$

Hence, there are two solutions in $[0,\ 2\pi].$

1. $\sin17\text{x}-\sin11\text{x}$

Solution:

We have,

$2(1-2\sin^27​​\text{x})\sin3\text{x}=2(\cos14\text{x})\sin3\text{x}$ $[\because\cos2\text{x}=1-2\sin^2\text{x}]$

$=2\sin3\text{x}\cos14\text{x}$

$=\sin17\text{x}-\sin11\text{x}$ $[\because2\sin\text{A}\cos\text{xB}=\sin(\text{A+B})-\sin(\text{A}-\text{B})]$

$\therefore2(1-2\sin^27\text{x})\sin3\text{x}=\sin17\text{x}-\sin11\text{x}$

2. $0$

Solution:$$

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

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

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

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

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

$=\ 0$

3. 6

Solution:

Given:

$3\sin^2\text{x}-7\sin\text{x}+2=0$

$\Rightarrow3\sin^2\text{x}-6\sin\text{x}-\sin\text{x}+2=0$

$\Rightarrow3\sin\text{x}(\sin\text{x}-2)-1(\sin\text{x}-2)=0$

$\Rightarrow(3\sin\text{x}-1)(\sin\text{x}-2)=0$

$\Rightarrow3\sin\text{x}-1=0$ or $\sin\text{x}-2=0$

Now, $\sin\text{x}=2$ is not possible, as the value of sin xsin x lies between -1 and 1.

$\Rightarrow\sin\text{x}=\frac{1}{4}$

Also, sin x is positive only in first two quadrants. Therefore, sin x is positive twice in the interval $[0,\ \pi].$

Hence, it is positive six times in the interval $[0,\ 5\pi]$ viz $[0,\ \pi],$ $2\pi,\ 3\pi$ and $[4\pi,\ 5\pi].$

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

Solution:

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

$=\cos^2\text{x} +\sec^2\text{x}-2\cos\text{x}\sec\text{x}+2\cos\text{x}\sec\text{x}$

$=(\sec\text{x}-\cos\text{x})^2 +2$

$\therefore\text{f}\text({x})\geq 2 \ \forall \text{ x }$$\Big[(\sec\text{x}-\cos\text{x})^2\geq 0 \ \forall \text{ x}\Big]$

Hence, the correct option is answer D.

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

Solution:

Given:

$7\cos^2\text{x}+3\sin^2\text{x}=4$

$\Rightarrow7\cos^2\text{x}+3\Big(1-\cos^2\text{x}\Big)=4$

$\Rightarrow7\cos^2\text{x}+3-3\cos^2\text{x}=4$

$\Rightarrow4\cos^2\text{x}+3=4$

$\Rightarrow4\Big(1-\cos^2\text{x}\Big)=3$

$\Rightarrow4\sin^2\text{x}=3$

$\Rightarrow\sin^2\text{x}=\frac{3}{4}$

$\Rightarrow\sin\text{x}=\frac{\sqrt{3}}{2}$

$\Rightarrow\sin\text{x}=\sin\frac{\pi}{3}$

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

2. $\sin\text{x}$

Solution:

We have,

$\frac{\sin3\text{x}}{1+2\cos2\text{x}}=\frac{3\sin\text{x}-4\sin^2\text{x}}{1+2(1-2\sin^2\text{x})}$

$=\frac{3\sin\text{x}-4\sin^3\text{x}}{1+2-4\sin\text{x}}$

$=\frac{\sin\text{x}(3-4\sin^2\text{x})}{(3-4\sin^2\text{x})}$

$=\sin\text{x}$

4. 2

Solution:

Given equation:

$\tan\text{x}+\tan2\text{x}=-\tan3\text{x}+\tan\text{x}\tan2\text{x}\tan3\text{x}$

$\Rightarrow\tan\text{x}+\tan2\text{x}=-\tan3\text{x}+\tan\text{x}\tan2\text{x}\tan3\text{x}$

$\Rightarrow\tan\text{x}+\tan2\text{x}=-\tan3\text{x}\Big(1-\tan\text{x}\tan2\text{x}\Big)$

$\Rightarrow\frac{\tan\text{x}+\tan2\text{x}}{1-\tan\text{x}\tan2\text{x}}=-\tan3\text{x}$

$\Rightarrow\tan(\text{x}+2\text{x})=-\tan3\text{x}$

$\Rightarrow\tan3\text{x}=-\tan3\text{x}$

$\Rightarrow2\tan3\text{x}=0$

$\Rightarrow\tan3\text{x}=0$

$\Rightarrow3\text{x}=\text{n}\pi$

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

Now,

$\text{x}=\frac{\pi}{3},\ \text{n}=1$

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

$\text{x}=\frac{3\pi}{3}=180^\circ,$ which is not possible, as it is not in the interval $(0,\ 2\pi).$

Hence, the number of solutions of the given equation is 2.

2. $\frac32$

Solution:

$\frac{2\pi}{3}=120^\circ$

Let $\text{f(x)}=\sin^2(90+30+\text{x})+\sin^2(90+30-\text{x})$

$=[\cos(30+\text{x})]^2+[\cos(30=\text{x})]^2$ $[\text{Using }\sin(90+\text{A})=\cos\text{A}]$

$=\Big[\frac{\sqrt{3}}{2}\cos\text{x}-\frac12\sin\text{x}\Big]^2+\Big[\frac{\sqrt{3}}{2}\cos\text{x}-\frac12\sin\text{x}\Big]^2$

$=\frac{\sqrt{3}}{2}\cos^2\text{x}-\frac14\sin^2\text{x}-\frac{\sqrt{3}}{2}\cos\text{x}\sin\text{x}+\frac34\cos^2\text{x}+\frac14\sin^2\text{x}+\frac{\sqrt{3}}{2}\cos\text{x}\sin\text{x}$

$=\frac{3}{2}\cos^2\text{x}-\frac12\sin^2\text{x}$

$=\frac{3}{2}(1-\sin^2\text{x})+\frac12\sin^2\text{x}$

$=\frac32-\frac{3}{2}\sin^2\text{x}+\frac12\sin^2\text{x}$

$=\frac32-\sin^2\text{x}$.

For f(x) to be maximum, $\sin^2\text{x}$ must have minimum value, which is 0.

$\therefore\frac32$ is the maximum value of f(x).

1. $\frac{\sqrt{5}+1}{8}$

Solution:

$\cos^248^\circ-\sin^212^\circ$

$=\cos(48^\circ+12^\circ)\cos(48^\circ-12^\circ)$ $[\cos(\text{A+B})\cos(\text{A}-\text{B})=\cos^2\text{A}-\sin^2\text{B}]$

$=\cos60^\circ\cos36^\circ$

$=\frac{1}{2}\times\Big(\frac{\sqrt{5}+1}{4}\Big)$

$=\frac{\sqrt{5}+1}{8}$

Hence, the correct answer is option A.

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. 

4. None of these

Solution:

We have,

$\tan\theta\sin\big(\frac{\pi}{2}+\text{x}\cos\big)\big(\frac{\pi}{2}-\text{x}\big)$

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

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

4. None of these

Solution:

Given equation: $\text{e}^{\sin\text{x}}-\text{e}^{-\sin\text{x}}-4=0,$

Let:

$\text{e}^{\sin\text{x}}=\text{y}$

Now,

$\text{y}-\text{y}^{-1}-4=0$

$\Rightarrow\text{y}^2-4\text{y}-1=0$

$\therefore\text{y}=\frac{4\pm\sqrt{16+4}}{2}$

$\Rightarrow\text{y}=\frac{4\pm\sqrt{20}}{2}$

$\Rightarrow\text{y}=\frac{4\pm2\sqrt{5}}{2}=2\pm\sqrt{5}$

And

$\text{y}=\text{e}^{\sin\text{x}}$

$\Rightarrow\text{y}^{\sin\text{x}}={2\pm\sqrt{5}}$

Taking log on both sides, we get:

$\sin\text{x}=\log_\text{e}\big(2\pm\sqrt{5}\big)$

$\Rightarrow\sin\text{x}=\log_{e}\big(2+\sqrt{5}\big)$ or $\sin\text{x}=\log_\text{e}\big(2-\sqrt{5}\big)$

$\Rightarrow\sin\text{x}=\log_{e}\big(4.24\big)$ or $\sin\text{x}=\log_\text{e}\big(-0.24\big)$

$\log(4.24)>1$ and $\sin\text{x}$ cannot be greater than 1.

In the other case, the log of negative term occurs, which is not defined.

3. 0

Solution:

Given that, $\sin\theta+\cos\theta=1$

Squaring both sides, we get,

$\Rightarrow(\sin\theta+\cos\theta)^2=(1)^2$ 

$\Rightarrow\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=1$

$\Rightarrow1+\sin2\theta=1$

$\Rightarrow\sin2\theta=1-1=0$

Hence, the correct option is (c).

1. $\cos\text{x}$

Solution:

We have,

$\therefore\frac{\cos3\text{x}}{2\cos2\text{x}-1}=\frac{4\cos^3​\text{x}-3\cos​\text{x}}{1(2\cos^2​\text{x}-1)-1}$ $[\therefore\cos3​\text{x}=4\cos^3​\text{x}-3\cos​\text{x}]$

$=\frac{4\cos^3​\text{x}-3\cos​\text{x}}{4\cos^2​\text{x}-2-1}$

$=\frac{4\cos^3​\text{x}-3\cos​\text{x}}{4\cos^2​\text{x}-3}$

$=\cos​\text{x}\Big(\frac{4\cos^2​\text{x}-3}{4\cos^2​\text{x}-3}\Big)$

$=\cos​\text{x}$