MATHS 3 Marks Question

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

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

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

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

Solution:

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

$=-\tan45^\circ$

$=-1$

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

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

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

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

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

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. $1$

Solution:

$\pi=180^\circ$

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

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

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

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

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

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

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

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

$=1$

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. $\sqrt{3}$

Solution:.

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

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

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

$=\tan60^\circ$

$=\sqrt{3}$

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).

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. $\frac\pi4$

Solution:

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

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

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

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

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

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

$=1$

$\therefore\alpha+\beta=\frac\pi4\ (\tan\frac\pi4=1)$

Hence, the correct answer is option D.

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,

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

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

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

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

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,

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

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

$=2$

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.

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. $\cos\text{A}\cos\text{B}=\frac15$

Solution:

$\tan\text{A}\tan\text{B}=\frac{\sin\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}=2\ (\text{Given})\cdots(1)$

Also,

$\cos(\text{A - B})=\frac35$

$\Rightarrow\cos\text{A}\cos\text{B}=\frac35+\sin\text{A}\sin\text{A}=\frac35$

$\therefore\sin\text{A}\sin\text{B}=\frac35-\cos\text{A}\cos\text{B}\cdots(2)$

Substituting eq (2) in eq (1), we get:

$\Rightarrow\frac{\frac35-\cos\text{A}\cos\text{B}}{\cos\text{A}\cos\text{B}}=2$

$\Rightarrow3\cos\text{A}\cos\text{B}=\frac35$

$\Rightarrow\cos\text{A}\cos\text{B}=\frac15$

1. $\tan55^\circ$

Solution:

$\frac{\cos10^\circ+\sin10^\circ}{\cos10^\circ-\sin10^\circ}$

$=\frac{1+\tan10^\circ}{1-\tan10^\circ}$ $[$Dividing the numerator and denominator by $\cos10^\circ]$

$=\frac{\tan45^\circ+\tan10^\circ}{1-\tan45^\circ\times\tan10^\circ}$

$=\tan(45^\circ+10^\circ)$ $\Big[$Using $\tan(\text{A+B})=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]$

$=\tan55^\circ$

1. $0$

Solution:.

$3\sin\text{x}+4\cos\text{x}=5$

$\frac{3}{5}\sin\text{x}+\frac{4}{5}\cos\text{x}=1$

Let $\cos\alpha=\frac35$ and $\sin\alpha=\frac45.$

$\therefore\cos\alpha\sin\text{x}+\sin\alpha\cos\text{x}=1$

$\Rightarrow\sin(\alpha+\text{x})=\sin\frac\pi2$

$\Rightarrow\alpha+\text{x}=\frac\pi2$

$\Rightarrow\text{x}=\frac{\pi}{2}-\alpha\cdots(1)$

We have to find the value of $4\sin\text{x}-3\cos\text{x}.$

$4\sin\Big(\frac\pi2-\alpha\Big)-3\cos\Big(\frac\pi3-\alpha\Big)$ ...{From eq (1)}

$=4\cos\alpha-3\sin\alpha$

$=4\times\frac35-3\times\frac45$ $\Big(\because\cos\alpha=\frac35$ and $\sin\alpha=\frac{4}{5}\Big)$

$=0$

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.

2. $\cos2\text{A}$

Solution:

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

$=\cos(36^\circ-\text{A})\cos(36^\circ+\text{A})+\sin[90^\circ-(54^\circ+\text{A})]\sin[90^\circ-(54^\circ-\text{A})]$

$=\cos(36^\circ-\text{A})\cos(36^\circ+\text{A})+\sin(36^\circ-\text{A})\sin(36^\circ-\text{A})$

$=\cos(36^\circ-\text{A}-36^\circ+\text{A})$$​​​​​​\big[\text{Using}\cos(\text{A - B) formula} \big]$

$=\cos2\text{A}$

3. a² - 2

Solution:

Given:

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

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

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

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

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

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

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

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

2. -1

Solution:

$\pi=180^\circ$

$\sec\text{A}(\cos\text{B}\cos\text{C}-\sin^2\text{B}\sin\text{C})=\frac{\cos\text{B}\cos(\pi-(\text{A+B}))-\sin\text{B}\sin(\pi(\text{A+B}))}{\cos\text{A}}$

We know that, $\cos(\pi-\theta)=-\cos\theta$ and $\sin(\pi-\theta)=\sin\theta,$

$\therefore\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{\cos\text{B}\cos(\text{A+B})-\sin\text{B}\sin(\text{A+B})}{\cos\text{A}}$

Now, using the identities $\cos(\text{A+B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$ and $\sin(\text{A+B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B},$ we get

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

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

$\Rightarrow\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{-\cos\text{A}}{\cos\text{A}}=-1$

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

Solution:.

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

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

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

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

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

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

Solution:

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

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

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

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

$\therefore$ value after 1st year $=600,000-90,000$

$=₹\ 510,000$

The equipment deprecites at the rate 135% in 2nd year i.e., $\frac{135}{1000}\times600,000=81000$

$\therefore$ value after 2 nd year =81000

The value after 3rd year $=\frac{12}{100}\times600000=72000$

The total depreciation in 10 years

$\Rightarrow\text{S}_{10}=\frac{10}{2}[2\times81000+(9)(-9000)]$

$=5[81000]$ $\big[$ using $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$

$\therefore$ The cost of machine aftar 10 years $=₹\ 600000-405000$

$=105000$

Then, S_n = 5n + 4 and s'_n 9n +16 respectively.

Then, if ratio of sum of n terms of 2A.P is giben, then the ratio of there n^th ther is obtained by replacing n by (2n - 1).

$\frac{\text{a}_\text{n}}{\text{a}_\text{n}}=\frac{5(2\text{n}-1)+1}{9(2\text{n}-1)+16}$

$\therefore$ Ratio of there 18th term is

$\frac{\text{a}_{18}}{\text{a}_{18}}=\frac{5(2\times18-1)+4}{9(2\times18-1)+16}$

$=\frac{5\times35+4}{9\times35+16}$

$=\frac{179}{321}$ 

$\text{d}=\text{x}$

$\text{a}_1=275$

$\text{S}=8000$

$\frac{16}{2}[\text{a}_1+\text{a}_1+(16-1)\times]=8000$

$2\text{a}_1+15\text{x=1000}$

$15\text{x}=450$

$\text{x}=30$

$=\text{a}_{16}$

$=\text{a}_1+(16-1)\text{d}$

$=275+15\times30$

$=725$