If sin α + sin β=a and cos α+cos β=b,

Question: If $\sin \alpha+\sin \beta=a$ and $\cos \alpha+\cos \beta=b$, prove that (i) $\sin (\alpha+\beta)=\frac{2 a b}{a^{2}+b^{2}}$ (ii) $\cos (\alpha-\beta)=\frac{a^{2}+b^{2}-2}{2}$ Solution: The given equations are $\sin \alpha+\sin \beta=a$ and $\cos \alpha+\cos \beta=b$. (i) $\because \sin C+\sin D=2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$ $\therefore 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=a \quad \ldots(1)$ Now, using the identity $\sin C+\sin D=2 \sin \frac{C+D}{2} \co...

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If sin α + sin β=a and cos α+cos β=b,

Question: If $\sin \alpha+\sin \beta=a$ and $\cos \alpha+\cos \beta=b$, prove that (i) $\sin (\alpha+\beta)=\frac{2 a b}{a^{2}+b^{2}}$ (ii) $\cos (\alpha-\beta)=\frac{a^{2}+b^{2}-2}{2}$ Solution: The given equations are $\sin \alpha+\sin \beta=a$ and $\cos \alpha+\cos \beta=b$. (i) $\because \sin C+\sin D=2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$ $\therefore 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=a \quad \ldots(1)$ Now, using the identity $\sin C+\sin D=2 \sin \frac{C+D}{2} \co...

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If sin α + sin β=a and cos α+cos β=b,

Question: If $\sin \alpha+\sin \beta=a$ and $\cos \alpha+\cos \beta=b$, prove that (i) $\sin (\alpha+\beta)=\frac{2 a b}{a^{2}+b^{2}}$ (ii) $\cos (\alpha-\beta)=\frac{a^{2}+b^{2}-2}{2}$ Solution: The given equations are $\sin \alpha+\sin \beta=a$ and $\cos \alpha+\cos \beta=b$. (i) $\because \sin C+\sin D=2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$ $\therefore 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=a \quad \ldots(1)$ Now, using the identity $\sin C+\sin D=2 \sin \frac{C+D}{2} \co...

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Find the shortest distance between the lines whose vector equations are

Question: Find the shortest distance between the lines whose vector equations are $\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})$ and $\vec{r}=4 \hat{i}+5 \hat{j}+6 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$ Solution: The given lines are $\vec{r}=\hat{i}+2 \hat{j}+3 \hat{k}+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})$ and $\vec{r}=4 \hat{i}+5 \hat{j}+6 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$ It is known that the shortest distance between the lines, $\vec{r}=\vec{a}_{1}+\la...

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The mean of the following data is 20.6 .Find the value of p.

Question: The mean of the following data is 20.6 .Find the value of p. Solution: It is given that, Mean = 20.6 $\Rightarrow \frac{25 p+530}{50}=20.6$ ⇒ 25p + 530 = 20.6 50 ⇒ 25p = 1030 530 ⇒ 25p = 500 ⇒ p = 500/25= 20 ⇒ p = 20 p = 20....

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If 2 tan α=3 tan β,

Question: If $2 \tan \alpha=3 \tan \beta$, prove that $\tan (\alpha-\beta)=\frac{\sin 2 \beta}{5-\cos 2 \beta}$ Solution: Given: $2 \tan \alpha=3 \tan \beta$ $\mathrm{LHS}=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \times \tan \beta}$ $=\frac{\frac{\frac{3}{2} \times \tan \beta-\tan \beta}{1+\frac{3}{2} \tan ^{2} \beta}}{(\because 2 \tan \alpha=3 \tan \beta)}$ $=\frac{\frac{1}{2} \times \tan \beta}{1+\frac{3}{2} \tan ^{2} \beta}=\frac{\tan \beta}{2+3 \tan ^{2} \beta}$ $=\frac{\frac{1}{2} \times...

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If sin θ=ab, find sec θ + tan θ in terms of a and b.

Question: If $\sin \theta=\frac{a}{b}$, find $\sec \theta+\tan \theta$ in terms of $a$ and $b$. Solution: Given: $\sin \theta=\frac{a}{b}$..... (1) To find: $\sec \theta+\tan \theta$ Now we know, $\sin \theta$ is defined as follows $\sin \theta=\frac{\text { Perpendicular side opposite to } \angle \theta}{\text { Hypotenuse }}$......(2) Now by comparing (1) and (2) We get, Perpendicular side opposite to $\angle \theta=\mathrm{a}$ and Hypotenuse = b Therefore triangle representing angle $\theta$ ...

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Find the shortest distance between the lines

Question: Find the shortest distance between the lines $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$ Solution: The given lines are $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$ It is known that the shortest distance between the two lines, $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ and $\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}$, is given by, $d=\frac{\l...

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Prove that:

Question: Prove that: $\cos \frac{\pi}{65} \cos \frac{2 \pi}{65} \cos \frac{4 \pi}{65} \cos \frac{8 \pi}{65} \cos \frac{16 \pi}{65} \cos \frac{32 \pi}{65}=\frac{1}{64}$ Solution: $\mathrm{LHS}=\cos \frac{\pi}{65} \cos \frac{2 \pi}{65} \cos \frac{4 \pi}{65} \cos \frac{8 \pi}{65} \cos \frac{16 \pi}{65} \cos \frac{32 \pi}{65}$ On dividing and multiplying by $2 \sin \frac{\pi}{65}$, we get $=\frac{1}{2 \sin \frac{\pi}{65}} \times 2 \sin \frac{\pi}{65} \times \cos \frac{\pi}{65} \times \cos \frac{2 \...

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Find the mean of the following data:

Question: Find the mean of the following data: Solution: $\therefore$ Mean $\overline{\mathrm{x}}=\frac{\sum \mathrm{fx}}{\mathrm{N}}$ = 2650/106 = 25....

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Calculate the mean for the following distribution:

Question: Calculate the mean for the following distribution: Solution: $\therefore$ Mean $\overline{\mathrm{x}}=\frac{\sum \mathrm{fx}}{\mathrm{N}}$ = 281/40 = 7.025....

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Prove that:

Question: Prove that: $\cos \frac{\pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}=\frac{-1}{16}$ Solution: $\cos \frac{\pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}$ $=\frac{1}{2 \sin \frac{\pi}{5}}\left(2 \sin \frac{\pi}{5} \cos \frac{\pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}\right) \quad$ (Multiplying and dividing by $\frac{1}{2 \sin \frac{\pi}{5}}$ ) $=\frac{1}{2 \sin \frac{\pi}{2}}\left(\sin \frac{2 \pi}{5} \cos \fr...

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Prove that:

Question: Prove that: $\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}=\frac{1}{16}$ Solution: $\mathrm{LHS}=\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}$ On dividing and multiplying by $2 \sin \frac{2 \pi}{15}$, we get $=\frac{1}{2 \sin \frac{2 \pi}{15}} \times\left(2 \sin \frac{2 \pi}{15} \times \cos \frac{2 \pi}{15}\right) \times \cos \frac{4 \pi}{15} \times \cos \frac{8 \pi}{15} \times \cos \frac{16 \pi}{15}...

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Find the shortest distance between the lines

Question: Find the shortest distance between the lines $\vec{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=2 \hat{i}-\hat{j}-\hat{k}+\mu(2 \hat{i}+\hat{j}+2 \hat{k})$ Solution: The equations of the given lines are $\vec{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ $\vec{r}=2 \hat{i}-\hat{j}-\hat{k}+\mu(2 \hat{i}+\hat{j}+2 \hat{k})$ It is known that the shortest distance between the lines, $\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1}$ and $\vec{r}=\...

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Ifis the mean of the ten natural numbers

Question: is the mean of the ten natural numbers $x_{1}, x_{2}, x_{3}, \cdots, x_{10}$ show that $\left(\mathrm{x}_{1}-\overline{\mathrm{X}}\right)+\left(\mathrm{x}_{2}-\overline{\mathrm{X}}\right)+\cdots+\left(\mathrm{x}_{10}-\overline{\mathrm{X}}\right)=0$ Solution: We have, $\overline{\mathrm{x}}=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\cdots+\mathrm{x}_{10}}{10}$ $\Rightarrow \mathrm{x}_{1}+\mathrm{x}_{2}+\cdots+\mathrm{x}_{10}=10 \overline{\mathrm{x}} \cdots \cdot(1)$ Now, $\left(\mathrm{x}_{1}...

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If tan θ=247, find that sin θ + cos θ.

Question: If $\tan \theta=\frac{24}{7}$, find that $\sin \theta+\cos \theta$ Solution: Given: $\tan \theta=\frac{24}{7}$....(1) To find: $\sin \theta+\cos \theta$ Now we know $\tan \theta$ is defined as follows $\tan \theta=\frac{\text { Perpendicular side opposite to } \angle \theta}{\text { Base side adjacent to } \angle \theta}$....(2) Now by comparing equation (1) and (2) We get Perpendicular side opposite to $\angle \theta=24$ Base side adjacent to $\angle \theta=7$ Therefore triangle repre...

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Find the sum of the deviations of the variate values

Question: Find the sum of the deviations of the variate values 3, 4, 6, 7, 8, 14 from their mean. Solution: Values 3, 4, 6, 7, 8, 14 $\therefore$ Mean $=\frac{\text { Sum of numbers }}{\text { Total numbers }}$ $\therefore$ Mean $=\frac{3+4+6+7+8+14}{6}$ Mean = 42/6 = 7 Sum of deviation of values from their mean =(3 7) + (4 7) + (6 7) + (7 7) + (8 7) + (14 7) = 4 3 1 + 0 + 1 + 7 = 8 + 8 = 0...

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Prove that:

Question: Prove that: $\cos 7^{\circ} \cos 14^{\circ} \cos 28^{\circ} \cos 56^{\circ}=\frac{\sin 68^{\circ}}{16 \cos 83^{\circ}}$ Solution: LHS $=\cos 7^{\circ} \cos 14^{\circ} \cos 28^{\circ} \cos 56^{\circ}$ On dividing and multiplying by $2 \sin 7^{\circ}$, we get $=\frac{1}{2 \sin 7^{\circ}} \times 2 \sin 7^{\circ} \times \cos 7^{\circ} \times \cos 14^{\circ} \times \cos 28^{\circ} \times \cos 56^{\circ}$ $=\frac{2 \sin 14^{\circ}}{2 \times 2 \sin 7^{\circ}} \times \cos 14^{\circ} \times \co...

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Show that the lines

Question: Show that the lines $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are perpendicular to each other. Solution: The equations of the given lines are $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ The direction ratios of the given lines are 7, 5, 1 and 1, 2, 3 respectively. Two lines with direction ratios, $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$, are perpendicular to each other, if $a_{1} a_{2}+b_{1} b_{2}...

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The sum of the deviations of a set of n values

Question: The sum of the deviations of a set of $n$ values $x_{1}, x_{2}, x_{3}, \cdots, x_{n}$ measured from 15 and $-3$ are $-90$ and 54 respectively. Find the value of $n$ and mean. Solution: Given: $\sum_{\mathrm{n}}^{\mathrm{i}=1}\left(\mathrm{x}_{\mathrm{i}}-15\right)=-90$ $\Rightarrow\left(x_{1}-15\right)+\left(x_{2}-15\right)+\cdots \cdot \cdot+\left(x_{n}-15\right)=-90$ $\Rightarrow\left(x_{1}+x_{2}+\cdots+n\right)-(15+15+15+\cdots \cdots+15)=-90$ $\Rightarrow \sum x-15 n=-90 \cdots(1)$...

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The sum of the deviations of a set of n values

Question: The sum of the deviations of a set of $n$ values $x_{1}, x_{2}, x_{3}, \cdots, x_{n}$ measured from 15 and $-3$ are $-90$ and 54 respectively. Find the value of $n$ and mean. Solution: Given: $\sum_{\mathrm{n}}^{\mathrm{i}=1}\left(\mathrm{x}_{\mathrm{i}}-15\right)=-90$ $\Rightarrow\left(x_{1}-15\right)+\left(x_{2}-15\right)+\cdots \cdot \cdot+\left(x_{n}-15\right)=-90$ $\Rightarrow\left(x_{1}+x_{2}+\cdots+n\right)-(15+15+15+\cdots \cdots+15)=-90$ $\Rightarrow \sum x-15 n=-90 \cdots(1)$...

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If tan

Question: If $\tan A=\frac{1}{7}$ and $\tan B=\frac{1}{3}$, show that $\cos 2 A=\sin 4 B$ Solution: Given: $\tan A=\frac{1}{7}$ and $\tan B=\frac{1}{3}$ Using the identity $\tan 2 B=\frac{2 \tan B}{1-\tan ^{2} B}$, we get $\tan 2 B=\frac{2 \times \frac{1}{3}}{1-\frac{1}{9}}=\frac{3}{4}$ Now, using the identities $\cos 2 A=\frac{1-\tan ^{2} A}{1+\tan ^{2} A}$ and $\sin 4 B=\frac{2 \tan 2 B}{1+\tan ^{2} 2 B}$, we get $\cos 2 A=\frac{1-\left(\frac{1}{7}\right)^{2}}{1+\left(\frac{1}{7}\right)^{2}}$ ...

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If cot θ=34, prove that sec θ−cosec θsec θ+cosec θ−−−−−−−−−√=17√.

Question: If $\cot \theta=\frac{3}{4}$, prove that $\sqrt{\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta+\operatorname{cosec} \theta}}=\frac{1}{\sqrt{7}}$ Solution: Given: $\cot \theta=\frac{3}{4}$.....(1) To prove: $\sqrt{\frac{\sec \theta-\operatorname{cosec} \theta}{\sec \theta+\operatorname{cosec} \theta}}=\frac{1}{\sqrt{7}}$ Now we know $\tan \theta$ is defined as follows $\cot \theta=\frac{\text { Base side adjacent to } \angle \theta}{\text { Perpendicular side opposite to } \...

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If tan x

Question: If $\tan x=\frac{b}{a}$, then find the value of $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$.[NCERT] Solution: Given: $\tan x=\frac{b}{a}$ $\therefore \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$ $=\sqrt{\frac{1+\frac{b}{a}}{1-\frac{b}{a}}}+\sqrt{\frac{1-\frac{b}{a}}{1+\frac{b}{a}}}$ $=\sqrt{\frac{1+\tan x}{1-\tan x}}+\sqrt{\frac{1-\tan x}{1+\tan x}}$ $=\sqrt{\frac{1+\frac{\sin x}{\cos x}}{1-\frac{\sin x}{\cos x}}}+\sqrt{\frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}}$ $=...

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Find the values of p so the line

Question: Find the values of $p$ so the line $\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}$ and $\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles. Solution: The given equations can be written in the standard form as $\frac{x-1}{-3}=\frac{y-2}{\frac{2 p}{7}}=\frac{z-3}{2}$ and $\frac{x-1}{\frac{-3 p}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}$ The direction ratios of the lines are $-3, \frac{2 p}{7}, 2$ and $\frac{-3 p}{7}, 1,-5$ respectively. Two lines with direction ratios, $a_{1}, b_...

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