MATHS M.C.Q (1 Marks)

3. $-\frac{1}{\sqrt{5}}$

Solution:

Given that, $\sin\theta=-\frac{4}{5},\theta$ lies in third quadrant

$\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\Big(\frac{-4}{5}\Big)^2}$

$=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\pm\frac{3}{5}$

$\therefore\cos\theta=-\frac{3}{5},\theta$ lies in third quadrant

$\cos\theta=2\cos^2\frac{\theta}{2}-1\Big[\because\pi<\theta\frac{3\pi}{2},\therefore\frac{\pi}{2}<\frac{\theta}{2}<\frac{3\pi}{4}\Big]$

$\Rightarrow\frac{-3}{5}=2\cos^2\frac{\theta}{2}-1$

$\Rightarrow2\cos^2\frac{\theta}{2}=1-\frac{3}{5}=\frac{2}{5}\Rightarrow\cos^2\frac{\theta}{2}=\frac{2}{5\times2}=\frac{1}{5}$

$\Rightarrow\cos\frac{\theta}{2}=\pm\frac{1}{\sqrt5}$

$\Rightarrow\cos\frac{\theta}{2}=-\frac{1}{\sqrt5}\Big[\because\frac{\pi}{2}<\frac{\theta}{2}<\frac{3\pi}{4}\Big]$

Hence, the correct option is (c).

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

Solution:

Given: $\text{(c + a + b) (a + b − c) = ab}$

$\Rightarrow(\text{a + b})^2-\text{c}^2=\text{ab}$

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

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

$\Rightarrow\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}=-\frac{1}{2}$

$\Rightarrow\cos\text{C}=-\frac{1}{2}=\cos\frac{2\pi}{3}$ (Using cosine rule)

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

Thus, the measure of angle C is $\frac{2\pi}{3}.$

Hence, the correct answer is option (c).

2. $\sin1^\circ<\sin1$

Solution:

We know that, 1 radian is approximately $57^\circ$.

Also, the value of $\sin\text{x}$ is always increasing for $0\leq \text{x}\leq 90^\circ$ 

$($or $\sin\text{x}$ is an increasing function for $0\leq \text{x}\leq 90^\circ).$

Now, $1^\circ < 57^\circ$

or $1^\circ< 1 \text{ radian}$

$\therefore\sin 1^\circ < \sin1$

Hence, the correct answer is option B.

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

Solution:

Let $\triangle\text{ABC}$ be the given triangle such that its sides are in the ratio $1:\sqrt{3}:2.$

$\therefore\text{a = k,b}=\sqrt{3}\text{k,c}=2\text{k}$

Now, $\text{a}^2+\text{b}^2=\text{k}^2+3\text{k}^2=4\text{k}^2=\text{c}^2$

So, $\triangle\text{ABC}$ is a right triangle right angled at C.

$\therefore\text{C}=90^{\circ}$

Using sine rule, we have

$\frac{\text{a}}{\sin\text{A}}=\frac{\text{b}}{\sin\text{B}}=\frac{\text{c}}{\sin\text{C}}$

$\Rightarrow\frac{\text{k}}{\sin\text{A}}=\frac{\sqrt{3}\text{k}}{\sin\text{B}}=\frac{2\text{k}}{\sin90^{\circ}}$

$\Rightarrow\sin\text{A}=\frac{1}{2}$ and $\sin\text{B}=\frac{\sqrt{3}}{2}$

$\Rightarrow\text{A}=30^{\circ}$ and $\text{B}=60^{\circ}$

Thus, the measure of its greatest angle is $\frac{\pi}{2}$

Hence, the correct answer is option (c).

2. $20\text{cm}$

Solution:

$\theta=\frac{\text{Arc}}{\text{Radius}}$

$\Rightarrow\frac{3\pi}{4}=\frac{15\pi}{\text{Radius}}$

$\Rightarrow\frac{60}{3}$

$\Rightarrow20\text{cm}$ 

3. $\cos4\text{x}$

Solution:

$\cos^4​​\text{x}+\sin^4\text{x}-6\cos^2\text{x}\sin^2\text{x}=\cos^4\text{x}\\+\sin^4\text{x}-2\cos^2\text{x}\sin^2\text{x}-4\cos^2\text{x}\sin^2\text{x}$

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

$=\cos^22\text{x}-\sin^22\text{x}$

$=\cos4\text{x}$

2. $\frac{2\text{k}}{\text{x}^2+1}$

Solution:

We have:

$\sec\text{x} +\tan\text{x} = \text{k}\cdots(1)$

$\Rightarrow\frac{1}{\sec\text{x} + \tan\text{x}}=\frac{1}{\text{k}}$

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

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

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

Adding (1) and (2):

$2\sec\text{x}= \text{k} + \frac{1}{\text{k}}$

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

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

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

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

1. 1

Solution:

Given:

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

$\Rightarrow\frac{1}{\tan\text{x}}-\tan\text{x}=\text{a}$

$\Rightarrow1-\tan^2\text{x}=\text{a}\tan\text{x}$

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

It is a quadratic equation.

If $\tan\text{x}=\text{z},$ then the equation becomes

$\text{z}^2+\text{az}-1=0$

$\Rightarrow\text{z}=\frac{-\text{a}\pm\sqrt{\text{a}^2+4}}{2}$

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

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

There are two roots of the given equation, but we need to find the number of roots in the first quadrant.

There is exactly one root of the equation, that is, $\text{x}=\tan^{-1}\Big(\frac{-\text{a}+\sqrt{\text{a}^2+4}}{2}\Big).$

4. 2

Solution:

Given:

$\tan\big(\frac{\pi}{4}+\text{x}+\tan\big(\frac{\pi}{4}-\text{x}\big)=\lambda\sec2\text{x}$

$\Rightarrow\frac{\tan\frac{\pi}{4}+\tan\text{x}}{1-\tan\frac{\pi}{4}\times\tan\text{x}}+\frac{\tan\frac{\pi}{4}-\tan\text{x}}{1+\tan\frac{\pi}{4}\times\tan\text{x}}=\lambda2\text{x}$

$\Rightarrow\frac{1+\tan\text{x}}{1-\tan\text{x}}+\frac{1-\tan\text{x}}{1+\tan\text{x}}=\lambda\sec2\text{x}$

$\Rightarrow\frac{(1+\tan\text{x})^2+(1-\tan\text{x})^2}{(1-\tan\text{x})(1+\tan\text{x})}=\lambda\sec2\text{x}$

$\Rightarrow\frac{2(1+\tan^2\text{x})}{1-\tan^2\text{x}}=\lambda\sec2\text{x}$

$\Rightarrow\frac{2\sec^2\text{x}}{1-\tan^2\text{x}}=\lambda\sec2\text{x}$

$\Rightarrow\frac{2}{\cos^2(1-\tan^2\text{x})}=\lambda\sec2\text{x}$

$\Rightarrow\frac{2}{\cos^2\text{x}\Big(1-\frac{\sin^2\text{x}}{\cos^2\text{x}}\Big)}=\lambda\sec2\text{x}$

$\Rightarrow\frac{2}{\cos^2\text{x}-\sin^2\text{x}}=\lambda\sec2\text{x}$

$\Rightarrow\frac{2}{\cos2\text{x}}=\lambda\sec2\text{x}$

$\Rightarrow2\sec2\text{x}=\lambda\sec2\text{x}$

$\Rightarrow2=\lambda$

$\therefore\lambda=2$

3. 22 : 13

Solution:

Let the angle subtended at the by the arec and radii of the first second circle $\theta_{1}$ and $\text{r}_{1}$ and $\theta_{2}$ and $\text{r}_{2}.$ 

We have,

$\theta_{1}=65^{\circ}=\Big(65\times\frac{\pi}{180}\Big)\ \text{radian}$

$\theta_{2}=65^{\circ}=\Big(110\times\frac{\pi}{180}\Big)\ \text{radian}$

$\theta_{1}=\frac{1}{\text{r}_{1}}$

$\Rightarrow \text{r}_{1}=\frac{1}{\big(65\times\frac{\pi}{180}\big)}$

$\Rightarrow \text{r}_{2}=\frac{1}{\big(110\times\frac{\pi}{180}\big)}$

$\frac{\text{r}_{1}}{\text{r}_{2}}=\frac{\frac{l}{\big(65\times\frac{\pi}{180}\big)}}{\frac{i}{\big(110\times\frac{\pi}{180}\big)}} $

$=\frac{110}{65}=\frac{22}{13}$

$\text{r}_{1}:\text{r}_{2}=22:13$​​​​​​​

1. $\sin\alpha$

Solution:

Given:

$\cot(\alpha+\beta)=0$

$\Rightarrow\frac{\cos(\alpha+\beta)}{\sin(\alpha+\beta)}=0$

$\Rightarrow\cos(\alpha+\beta)=0$

$\Rightarrow\alpha+\beta=\frac\pi2$

$\therefore\sin(\alpha+2\beta)=\sin(\alpha+\alpha+\beta)$

$=\sin\alpha$

3. $4\tan\beta$

Solution:

We have,

$5\sin\alpha=3\sin(\alpha+2\beta)$

$\Rightarrow\frac{5}{3}=\frac{\sin(\alpha+2\beta)}{\sin\alpha}$

$\Rightarrow\frac{5-3}{5+3}=\frac{\sin(\alpha+2\beta)-\sin\alpha}{\sin(\alpha+2\beta)+\sin\alpha}$ (using componendo and dividendo)

$\Rightarrow\frac{2}{8}=\frac{\sin(\alpha+2\beta)-\sin\alpha}{\sin(\alpha+2\beta)+\sin\alpha}$

$\Rightarrow\frac{1}{4}=\frac{2\cos\frac{\alpha+2\beta+\alpha}{2}\sin\frac{\alpha+2\beta-\alpha}{2}}{2\sin\frac{\alpha+2\beta+\alpha}{2}\cos\frac{\alpha+2\beta-\alpha}{2}}$

$\Rightarrow\frac{1}{4}\frac{\cos(\alpha+\beta)\sin\beta}{\sin(\alpha+\beta)\cos\beta}$

$\Rightarrow\frac{1}{4}=\cot(\alpha+\beta)\tan\beta$

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

$\therefore\tan(\alpha+\beta)=4\tan\beta$

2. b

Solution:

Given that, $\tan\theta=\frac{\text{a}}{\text{b}}$ 

$\text{b}\cos2\theta+\text{a}\sin2\theta=\text{b}\Big[\frac{1-\tan^2\theta}{1+\tan^2\theta}\Big]+\text{a}\Big[\frac{2\tan\theta}{1+\tan^2\theta}\Big]$

$=\text{b}\Bigg[\frac{1-\frac{\text{a}^2}{\text{b}^2}}{1+\frac{\text{a}^2}{\text{b}^2}}\Bigg]+\text{a}\Bigg[\frac{\frac{2\text{a}}{\text{b}}}{1+\frac{\text{a}^2}{\text{b}^2}}\Bigg]$

$=\text{b}\Big[\frac{\text{b}^2-\text{a}^2}{\text{b}^2+\text{a}^2}\Big]+\Bigg[\frac{\frac{2\text{a}^2}{\text{b}}}{\frac{\text{b}^2+\text{a}^2}{\text{b}^2}}\Bigg]$

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

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

Hence, the correct option is (b).

4. None of these

Solution:

Given equation:

$3\cos\text{x}+4\sin\text{x}=6\ .....(1)$

Thus, the equation is of the form $\text{a}\cos\text{x}+\text{b}\sin\text{x}=\text{c},$ where  and $\text{c}=6.$

Let:

$\text{a}=3=\text{r}\cos\alpha$ and $\text{b}=4=\text{r}\sin\alpha$

Now,

$\tan\alpha=\frac{\text{b}}{\text{a}}=\frac{4}{3}$

$\Rightarrow\alpha=\tan^{-1}\Big(\frac{4}{3}\Big)$

Also,

$\text{r}=\sqrt{\text{a}^2+\text{b}^2}=\sqrt{9+16}=\sqrt{25}=5$

On putting $\text{a}=3=\text{r}\cos\alpha$ and $\text{b}=4=\text{r}\sin\alpha$ in equation (i), we get:

$\Rightarrow\text{r}\cos(\theta-\alpha)=6$

$\Rightarrow\text{5}\cos(\theta-\alpha)=6$

$\Rightarrow\text{}\cos(\theta-\alpha)=\frac{6}{5}$

From here, we cannot find the value of $\theta.$

4. $8\pi$

Solution:

$\text{Angular velocity}=\frac{\text{Distance}}{\text{Time}}$

$=\frac{4\times2\pi}{1}$

$=8\pi\ \text{radians}$

1. $\frac{\sqrt{5}}{\sqrt{6}}$

Solution:

In the fourth quadrant, $\cos\text{x}\text{ and }\sec\text{x}$ are positive.

$\cos\text{x}=\frac{1}{\sec\text{x}}$

 $=\frac{1}{\sqrt{\sec^2\text{x}}}$

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

 $=\frac{1}{\sqrt{1+\Big(-\frac{1}{\sqrt{5}}\Big)^2}}$

$=\frac{1}{\sqrt{\frac{6}{5}}}$

$=\frac{\sqrt{5}}{\sqrt{6}}$

4. None of these

Solution:

ABC is a tringle.

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

$\Rightarrow\text{A}+\text{B}+\pi-\text{C}$

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

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

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

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

$\Rightarrow0=\tan\text{A}+\tan\text{B}\tan\text{C}$ $\big[$ given: $\tan\text{A}\tan\text{B}\tan\text{C}=0\big]$

$\Rightarrow\tan\text{A}\tan\text{B}\tan\text{C}=0$

$\Rightarrow\frac{1}{\tan\text{A}\tan\text{B}\tan\text{C}}=\frac{1}{0}$

$\Rightarrow\cot\text{A}\cot\text{B}\cot\text{C}\rightarrow\infty$

3. $1$

Solution:

$\pi=180^\circ$

Using $\tan(180^\circ-\text{A})=-\tan\text{A},$ we get:

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

Now, $\frac{\tan\text{A}+\tan\text{B}+\tan\text{C}}{\tan\text{A}\tan\text{B}\tan\text{C}}$

$=\frac{\tan\text{A}+\tan\text{B}-\tan[\pi-\text{(A+B)}]}{\tan\text{A}\tan\text{B}\tan[\pi-\text{(A+B)}]}$

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

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

$=\frac{\tan\text{A}+\tan\text{B}-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}-\tan\text{A}-\tan\text{B}}{-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}}$

$=\frac{-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}}{-\tan^2\text{A}\tan\text{B}-\tan\text{A}\tan^2\text{B}}$

$=1$

1. $16\cos^4-12\cos^2\text{x}+1$

Solution:

To find: $\frac{\sin5\text{x}}{\sin\text{x}}$

Now,

$\sin5\text{x}=\sin(3\text{x}+2\text{x})$

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

$=(3\sin\text{x}-4\sin^3\text{x}-4\sin^3\text{x}+8\sin^5\text{x})+2\sin\text{x}\cos^2\text{x}(4\cos^2\text{x}-3)$

$=(3\sin\text{x}-10\sin^3\text{x}+8\sin^5\text{x}+2\sin\text{x}(1-\sin^2\text{x})[2(1-\sin^2\text{x})-3]$

$=(3\sin\text{x}-10\sin^3\text{x}+8\sin^5\text{x}+(2\sin\text{x}-2\sin^3​​\text{x})(4-4\sin^2\text{x}-3)]$

$=(3\sin\text{x}-10\sin^3\text{x}+8\sin^5\text{x}+(2\sin\text{x}-8\sin^3​​\text{x}2\sin^3\text{x}+8\sin^5​​\text{x})]$

$=5\sin\text{x}-20\sin^3+16\sin^5\text{x}$

$\therefore\frac{\sin5\text{x}}{\sin\text{x}}=\frac{5\sin\text{x}-20\sin^3\text{x}+16^5\text{x}}{\sin\text{x}}$

$=5-20\sin^2\text{x}+16\sin^4\text{x}$

$=5-20(1-\cos^2\text{x})+16(1-\cos^2\text{x})^2$

$=5-20+20\cos^2\text{x}+16(1+\cos^4\text{x}-2\cos^4\text{x})$

$=5-20+20\cos^2\text{x}+16+16\cos^4\text{x}-32\cos^2\text{x}$

$=16\cos^4-12\cos^2\text{x}+1$

3. $\frac{\text{D}}{100}=\frac{\text{G}}{100}=\frac{2\text{R}}{\pi}$

Solution:

It is the relation between degree, grade and radian.

2. b² - c²

Solution:

Using cosine rule, we have

$\text{a(b}\cos\text{C}-\text{c}\cos\text{B})$

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

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

$=\frac{2\text{b}^2-2\text{c}^2}{2}$

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

Hence, the correct answer is option (b).

1. $\sqrt{2}\tan\beta$

Solution:

Given:

$\cos2\alpha=\frac{3\cos2\beta-1}{3-\cos2\beta}$

$\Rightarrow\frac{\cos2\alpha-1}{\cos2\alpha+1}=\frac{(3\cos2\beta-1)-(3-\cos2\beta)}{(3\cos2\beta-1)+(3-\cos2\beta)}$ (Using componendo and dividendo)

$\Rightarrow\frac{\cos2\alpha-1}{\cos2\alpha+1}=\frac{4\cos2\beta-4}{2\cos2\beta+2}$

$\Rightarrow​-​=\frac{1-\cos^2\alpha}{1+\cos2\alpha}=\frac{-4(1-\cos2\beta)}{2(1+\cos2\beta)}$

$\Rightarrow\frac{1-\cos2\alpha}{1+\cos2\alpha}=\frac{2(1-\cos2\beta)}{(1+\cos2\beta)}$

$\Rightarrow\frac{2\sin^2\alpha}{2\cos^2\alpha}=\frac{2(2\sin^2\beta)}{2\cos^2\beta}$

$\Rightarrow\tan^2\alpha=2\tan^2\beta$

$\therefore\tan\alpha=\sqrt{2}\tan\beta$

2. $\frac{19\pi}{24}$

Solution:

Given:

$\tan(\text{A - B})=1$ and $\sec(\text{A+B})=\frac{2}{\sqrt{3}}$

$\Rightarrow\text{A - B}=\frac{\pi}{4}\cdots(1)$ and $\text{A + B}=\frac{\pi}{4}\cdots(2)$

Adding these equations we get:

$2\text{A}=\frac{\pi}{4}+\frac\pi6$

$\Rightarrow\text{A}=\frac{5\pi}{24}$

$\Rightarrow$ Smallest possible value of $\text{B}=\pi-\frac{5\pi}{24}=\frac{19\pi}{24}.$ 

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

Solution:

$\sin\frac{\pi}{10}\sin\frac{13\pi}{10}=\sin\frac{\pi}{10}\sin\Big(\pi+\frac{3\pi}{10}\Big)=-\sin\frac{\pi}{10}\sin\frac{3\pi}{10}$

$=-\sin18^\circ\sin54^\circ=-\sin18^\circ\cos36^\circ$

$=-\Big(\frac{\sqrt5-1}{4}\Big)\Big(\frac{\sqrt5+1}{4}\Big)=\frac{1-5}{16}=-\frac{1}{4}$

4. $\frac{\sin4^{\text{n}-1}\alpha}{4^{\text{n}-1}\sin\alpha}$

Solution:

$\therefore\cos\alpha\cos2\alpha\cos4\alpha\ ...\cos2^{\text{n}-1}\alpha=\frac{\sin2^​\text{n}\alpha​}{2^\text{n}\sin\alpha}$

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

Solution:

$\frac{5\pi}{12}=75^\circ,\frac{\pi}{12}=15^\circ$

$\sin^275^\circ-\sin^215^\circ$

$=\sin^275^\circ-\cos^275^\circ$ $[\sin(90^\circ-\theta)=\cos\theta]$.

Now, $\sin75^\circ=\sin(45^\circ+30^\circ)$

$=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ$

$=\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times\frac12$

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

$\cos75^\circ=\cos(45^\circ+30^\circ)$

$=\cos45^\circ\cos30^\circ-\sin45^\circ\sin30^\circ$

$=\frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times\frac12$

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

Hence,

$\sin^275^\circ-\cos^275^\circ=\Big(\frac{\sqrt{3}+1}{2\sqrt{2}}\Big)^2-\Big(\frac{\sqrt{3}-1}{2\sqrt{2}}\Big)^2$

$=\frac{3+1+2\sqrt{3}-3-1+2\sqrt{3}}{8}$

$=\frac{4\sqrt{3}}{8}$

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

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

Solution:

We have, $\tan\theta=\frac{1}{2}$ and $\tan\phi=\frac{1}{3}$

$\tan(\theta+\phi)=\frac{\tan\theta+\tan\phi}{1-\tan\theta\cdot\tan\phi}=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot\frac{1}{3}}=\frac{\frac{5}{6}}{1-\frac{1}{6}}=1$

$\therefore\theta+\phi=\frac{\pi}{4}$

1. AP

Solution:

Given:

$\tan\text{px}-\tan\text{qx}=0$

$\Rightarrow\tan​​\text{px}=\tan\text{qx}$

$\Rightarrow\frac{\sin\text{px}}{\cos\text{px}}=\frac{\sin\text{qx}}{\cos\text{qx}}$

$\Rightarrow\sin\text{px}\cos\text{qx}=\sin\text{qx}\cos\text{px}$

$\Rightarrow\frac{1}{2}\Big[\sin\Big(\frac{\text{p+q}}{2}\Big)\text{x}+\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}\Big]\\=\frac{1}{2}\Big[\sin\Big(\frac{\text{q+p}}{2}\Big)\text{x}+\sin\Big(\frac{\text{q}-\text{p}}{2}\text{x}\Big)\Big]$

Now,

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

$\Rightarrow\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=\sin\Big(\frac{\text{q}-\text{p}}{2}\Big)\text{x}$

$\Rightarrow\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=-\sin\Big(\frac{\text{q}-\text{p}}{2}\Big)\text{x}$

$\Rightarrow2\sin\Big(\frac{​\text{p}-\text{q}​}{2}\Big)​\text{x}=0$

$\Rightarrow\sin\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=0$

$\Rightarrow\Big(\frac{\text{p}-\text{q}}{2}\Big)\text{x}=\text{n}\pi,\text{n}\in\text{Z}$

Now, on putting the value of n, we get:

$\text{n}=1,\text{x}=\frac{2\pi}{(\text{p}-\text{q})}=\text{a}_1$

$\text{n}=2,\text{x}=\frac{4\pi}{(\text{p}-\text{q})}=\text{a}_2$

$\text{n}=3,\text{x}=\frac{6\pi}{(\text{p}-\text{q})}=\text{a}_3$

$\text{n}=4,\text{x}=\frac{4\pi}{(\text{p}-\text{q})}=\text{a}_4$

And so on.

Also,

$\text{d}=\text{a}_2-\text{a}_1=\frac{4\pi}{(\text{p}-\text{q})}-\frac{2\pi}{(\text{p}-\text{q})}=\frac{2\pi}{(\text{p}-\text{q})}$

$\text{d}=\text{a}_3-\text{a}_2=\frac{6\pi}{(\text{p}-\text{q})}-\frac{4\pi}{(\text{p}-\text{q})}=\frac{2\pi}{(\text{p}-\text{q})}$

$\text{d}=\text{a}_4-\text{a}_3=\frac{8\pi}{(\text{p}-\text{q})}-\frac{6\pi}{(\text{p}-\text{q})}=\frac{2\pi}{(\text{p}-\text{q})}$

And so on.

Thus, xx forms a series in AP.

3. 2

Solution:

Given equation is $\tan\text{x}+\sec\text{x}=2\cos\text{x}$

$\Rightarrow\frac{\sin\text{x}}{\cos\text{x}}+\frac{1}{\cos\text{x}}=2\cos\text{x}$

$\Rightarrow1+\sin\text{x}=2\cos^2\text{x}\Rightarrow2\cos^2\text{x}-\sin\text{x}-1=0$

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

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

since, the equation is a quadratic equation in $\sin\text{x}$ so it will have 2 solutions. 

Hence, the correct option is (c)

3. $\sqrt{3}$

Solution:.

$\tan20^\circ+\tan40^\circ+\sqrt{3}\tan20^\circ\tan40^\circ$

$=\tan60^\circ(1-\tan20^\circ\tan40^\circ)+\tan60^\circ\tan20^\circ\tan40^\circ$

$=\tan60^\circ-\tan60^\circ\tan20^\circ\tan40^\circ+\tan60^\circ\tan20^\circ\tan40^\circ$

$=\tan60^\circ$

$=\sqrt{3}$

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

Solution:

Given:

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

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

$\Rightarrow1-\sin^2\text{x}+\sin\text{x}+1=0$

$\Rightarrow\sin^2\text{x}-\sin\text{x}-2=0$

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

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

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

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

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

$\sin\text{x}=2$ is not possible.

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

$\therefore\sin\text{x}=\sin\frac{3\pi}{2}$

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

The values of x lies in the third and fourth quadrants.

Hence, x lies in $\Big(\frac{5\pi}{4},\ \frac{7\pi}{4}\Big).$

3. $\pm\frac13$

Solution:

$\sin(\pi\cos\text{x})=\cos(\pi\sin\text{x})$

As we know that $\sin\text{x}=-\cos\Big(\frac\pi2+\text{x}\Big)$

$\Rightarrow-\cos\Big(\frac\pi2+\pi\cos\text{x}\Big)=\cos(\pi\sin\text{x})$

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

$\Rightarrow\pi\sin\text{x}-\pi\cos\text{x}=\frac12$

$\Rightarrow\sin\text{x}-\cos\text{x}=\frac12$

Squaring both sides we get,

$\Rightarrow\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x}=\frac14$

$\Rightarrow1+\sin2\text{x}=\frac{1}{4}$

$\Rightarrow\sin2\text{x}=\frac13$

$\therefore\sin2\text{x}=\pm\frac13$

4. 3

Solution:

The given expression is $3\cos\text{x}+4\sin\text{x}+8$

Let $\text{y}=3\cos\text{x}+4\sin\text{x}+8$

$\Rightarrow\text{y}-8=3\cos\text{x}+4\sin\text{x}$

Minimum value of $\text{y}-8=-\sqrt{(3)^2+(4)^2}$

$\Rightarrow\text{y}-8=-\sqrt{9+16}=-5$

$\Rightarrow\text{y}=8-5=3$

so, the minimum value of the given expression is 3.

Hence, the correct option is (d).

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

Solution:

$\tan2​\text{x}+\sec2\text{x}=\frac{2\tan\text{x}}{1-\tan^2\text{x}}+\frac{1+\tan^2\text{x}}{1-\tan^2\text{x}}$

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

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

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

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

$\frac{1+\text{t}}{1-\text{t}}$ $[\tan\text{x}=\text{t}]$ (given)

2. $\frac\pi3$

Solution:

$\cos\text{P}=\frac17,\cos\text{Q}=\frac{13}{14}$

$\therefore\sin\text{P}=\sqrt{1-\frac{1}{49}}=\frac{4\sqrt{3}}{7}$ and $\sin\text{Q}=\sqrt{1-\frac{169}{196}}=\frac{3\sqrt{3}}{14}$

Hence, $\tan\text{P}=4\sqrt{3},\tan\text{Q}=\frac{3\sqrt{3}}{13}$

$\cos(\text{P - Q})=\cos\text{P}\cos\text{Q}+\sin\text{P}\sin\text{Q}$

$=\frac{1}{7}\times\frac{13}{14}+\frac{4\sqrt{3}}{7}\times\frac{3\sqrt{3}}{14}$

$=\frac{13+36}{98}$

$=\frac{49}{98}$

$\therefore\cos(\text{P - Q})=\frac12$

$\Rightarrow\text{P - Q}=\cos^{-1}\frac12$

$\Rightarrow\text{P - Q}=60^\circ$

Hence, the correct answer is option B.

3. $\frac{13\pi^{\text{c}}}{18}$

Solution:

We know that the hour of a clock completes one rotation in 12 hours = 360° 

Angle traced by the hour hand in 12 hours = 360°

Now,

Angle traced by the hour hand in 8 hours 30 minutes, i.e.

We also know that the minute hand of a clock completes one rotation in 60 minutes.

Angle traced by the minute hade in 30 minutes $=\Big(\frac{360}{60}\times40^{\circ}\Big)=240^{\circ}$

Required angle between the two hands of the clock = 240° - 110° = 130°

Value of the angle (in radians) between the two hands of the clock $\Big(130\times\frac{\pi}{180}\Big)^{\text{c}}=\frac{13\pi^{\text{c}}}{18}$

1. $[-1,3]$

Solution:

$\text{A}=2\sin^2\text{x}-\cos2\text{x}$

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

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

$\therefore0\leq\sin^2\text{x}\leq1$

$\Rightarrow4\times0\leq4\times\sin^2\text{x}\leq4\times1$

$\Rightarrow0\leq4\sin^2\text{x}\leq4$

$\Rightarrow0-1\leq4\sin^2\text{x}-1\leq4-1$

$\Rightarrow-1\leq4\sin^2\text{x}-1\leq3$

$\Rightarrow-1\leq\text{A}\leq3$

$\Rightarrow\text{A}\in[-1,3]$

3. 3

Solution:

Given that, $\tan\text{A}=\frac{1}{2}$ and $\tan\text{B}=\frac{1}{3}$

$\tan2\text{A}=\frac{2\tan\text{A}}{1-\tan^2\text{A}}=\frac{2\times\frac{1}{2}}{1-\Big(\frac{1}{2}\Big)^2}$

$=\frac{1}{1-\frac{1}{4}}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$

so, $\tan2\text{A}=\frac{4}{3}$ and $\tan\text{B}=\frac{1}{3}$

$\tan(\text{2A+B})=\frac{\tan2\text{A}+\tan\text{B}}{1-\tan\text{A}.\tan\text{B}}=\frac{\frac{4}{3}+\frac{1}{3}}{1-\frac{4}{3}\times\frac{1}{3}}$

$=\frac{\frac{5}{3}}{\frac{9-4}{9}}=\frac{5}{3}\times\frac{9}{5}=3$

Hence, the correct option is (c).

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]$

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

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

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

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

3. $\sec\theta=\frac{1}{2}$

Solution:

$\sin\theta=-\frac{1}{5}$ is correct. $\because-1\leq\sin\theta\leq1$ 

so (a) is correct.

$\cos\theta=1$ is correct. $\because\cos0^\circ=1$

so (b) is correct.

$\sec\theta=-\frac{1}{2}\Rightarrow\cos\theta=2$ is not correct. $\because-1\leq\cos\theta\leq1$

Hence, (c) is not correct.

4. $\frac\pi4$

Solution:

It is given that $\tan\theta=\frac12$ and $\tan\phi=\frac13.$

Now,

$\tan(\theta+\phi)=\frac{\tan+\tan\phi}{1-\tan\theta\tan\phi}$

$=\frac{\frac12+\frac13}{1-\frac12\times\frac13}$

$=\frac{\frac{5}{6}}{\frac56}$

$=1$

$\therefore\theta+\phi=\frac\pi4$ $\Big(\tan\frac\pi4=1\Big)$

Hence, the correct answer is option D.

1. $0$

Solution:

We have,

$2\tan\frac{\pi}{10}+3\sec\frac{\pi}{10}-4\cos\frac{\pi}{10}$

$=2\tan18^\circ+3\sec18^\circ-4\cos18^\circ$

$=2\frac{\sin18^\circ}{\cos18^\circ}+3\times\frac{1}{\cos18^\circ}-4\cos18^\circ$

$=2\times\frac{\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{10+2\sqrt{5}}}{4}}+3\times\frac{1}{\frac{\sqrt{10+2\sqrt{5}}}{4}}=4\times\frac{\sqrt{10+2\sqrt{5}}}{4}$

$=2\times\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}+3\times\frac{4}{\sqrt{10+2\sqrt{5}}}-\sqrt{10+2\sqrt{5}}$

$=\frac{2\sqrt{5}-2+12-\Big(\sqrt{10+2\sqrt{5}}\Big)^2}{\Big(\sqrt{10+2\sqrt{5}}\Big)}$

$=\frac{2\sqrt{5}+10-10-2\sqrt{5}}{\Big(\sqrt{10+3\sqrt{5}}\Big)}$

$=0$

4. 194

Solution:

We have:

$\tan\text{A}+\cot\text{A}=4$

squaring both the sides:

$(\tan\text{A}+\cot\text{A})^2=4^2$

$\Rightarrow\tan^2\text{A}+\cot^2\text{A}+2(\tan\text{A})(\cot\text{A})=16$

$\Rightarrow\tan^2\text{A}+\cot^2\text{A}+2=16$

$\Rightarrow\tan^2\text{A}+\cot\text{A}=14$

squaring both the sides again:

$(\tan^2\text{A}+\cot^2\text{A})^2=14^2$

$\tan^4\text{A}+\cot^4\text{A}+2(\tan^2\text{A})(\cot^2\text{A})=196$

$\Rightarrow\tan^4\text{A}+\cot^4\text{A}+2=196$

$\Rightarrow\tan^4\text{A}+\cot^4\text{A}=194$

3. a² - 2

Solution:

Given:

$\tan\Big(\frac\pi4+\text{x}\Big)+\tan\Big(\frac\pi4-\text{x}\Big)=\text{a}$

$\Rightarrow\Big[\tan\Big(\frac\pi4+\text{x}\Big)+\tan\Big(\frac\pi4-\text{x}\Big)\Big]^2=\text{a}^2$

$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)+2\tan\Big(\frac\pi4-\text{x}\Big)\tan\Big(\frac\pi4-\text{x}\Big)=\text{a}^2$

$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2\tan\Big(\frac\pi4-\text{x}\Big)\tan\Big(\frac\pi4-\text{x}\Big)$

$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2\Big[\frac{\tan45^\circ-\tan\text{x}}{1+\tan45^\circ\tan\text{x}}\times\frac{\tan45^\circ+\tan\text{x}}{1+\tan45^\circ\tan\text{x}}\Big]$

$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2\Big[\frac{1^\circ-\tan\text{x}}{1+\tan\text{x}}\times\frac{1^\circ+\tan\text{x}}{1-\tan\text{x}}\Big]$

$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2\Big(\frac{1-\tan^2\text{x}}{1-\tan^2\text{x}}\Big)$

$\Rightarrow\tan^2\Big(\frac\pi4+\text{x}\Big)+\tan^2\Big(\frac\pi4-\text{x}\Big)\text{a}^2-2$

1. $\frac{\sin2\beta}{5-\cos2\beta}$

Solution:

Given:

$2\tan\alpha=3\tan\beta$

Now,

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

$=\frac{\frac{3}{2}\tan\beta-\tan\beta}{1+\Big(\frac{3}{2}\tan\beta\Big)\tan\beta}$

$=\frac{3\tan\beta-2\tan\beta}{2+3\tan^2\beta}$

$=\frac{\tan\beta}{2+3\tan^2\beta}$

$=\frac{\frac{\sin\beta}{\cos\beta}}{2+3\frac{\sin^2\beta}{\cos^2\beta}}$

$=\frac{\sin\beta\cos\beta}{2\cos\beta+3\sin\beta}$

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

$=\frac{2\sin\beta\cos\beta}{4+2-\cos2\beta}$

$=\frac{\sin2\beta}{4+1-cos2\beta}$

$\therefore\tan(\alpha+\beta)=\frac{\sin2\beta}{5-\cos2\beta}$

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

Solution:

Given:

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

$\cos3\text{x}=\lambda\Big(\text{a}^2+\frac{1}{a^3}\Big)$

Now,

$\cos^3\text{x}=\frac{1}{8}\Big[\text{a}^3+\frac{1}{\text{a}^2}+3\text{a}\frac{1}{\text{a}^3}+\text{a}\frac{1}{\text{a}}\Big]$

$\cos^3\text{x}=\frac{1}{8}\Big(​a​^3+\frac{1}{​a​^3}+32\cos​​​\text{x}​\Big)$ $[\because\cos\text{x}=\frac{1}{2}(\text{a}+\frac{1}{\text{a}})]$

$\Rightarrow\cos^3\text{x}=\frac{1}{8}\Big(\frac{\cos^3\text{x}}{\lambda}+6\cos\text{x}\Big)$

$\Rightarrow\cos^3\text{x}=\frac{1}{8}\Big(\frac{4\cos^3\text{x}-3\cos\text{x}}{\lambda}+6\cos\text{x}\Big)$

$\Rightarrow\cos^3\text{x}=\frac{4\cos^3\text{x}}{8\lambda}\frac{3\cos^3\text{x}}{8\lambda}+\frac{6\cos\text{x}}{8}$

On comparing the powers of cos3x on both sides, we get

$1=\frac{4}{8\lambda}$

$1=\frac{4}{8\lambda}$

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

2. $0$

4. $\frac{3}{4}\leq\text{A}\leq1$