sin 5x=5 cos
Question: $\sin 5 x=5 \cos ^{4} x \sin x-10 \cos ^{2} x \sin ^{3} x+\sin ^{5} x$ Solution: LHS $=\sin 5 \mathrm{x}$ $=\sin (3 x+2 x)$ $=\sin 3 x \times \cos 2 x+\cos 3 x \times \sin 2 x$ $=\left(3 \sin x-4 \sin ^{3} x\right)\left(2 \cos ^{2} x-1\right)+\left(4 \cos ^{3} x-3 \cos x\right) \times 2 \sin x \cos x$ $=-3 \sin x+4 \sin ^{3} x+6 \sin x \cos ^{2} x-8 \sin ^{3} x \cos ^{2} x+8 \sin x \cos ^{4} x-6 \sin x \cos ^{2} x$ $=8 \sin x \cos ^{4} x-8 \sin ^{3} x \cos ^{2} x-3 \sin x+4 \sin ^{3} x...
Read More →Find the Cartesian equation of the following planes:
Question: Find the Cartesian equation of the following planes: (a) $\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=2$ (b) $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-4 \hat{k})=1$ (c) $\vec{r} \cdot[(s-2 t) \hat{i}+(3-t) \hat{j}+(2 s+t) \hat{k}]=15$ Solution: (a)It is given that equation of the plane is $\vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=2$ ...(1) For any arbitrary point $P(x, y, z)$ on the plane, position vector $\vec{r}$ is given by, $\vec{r}=x \hat{i}+y \hat{j}-z \hat{k}$ Substituting the value of $\vec{...
Read More →cot x + cot
Question: $\cot x+\cot \left(\frac{\pi}{3}+x\right)+\cot \left(\frac{2 \pi}{3}+x\right)=3 \cot 3 x$ Solution: $\frac{\pi}{3}=60^{\circ}, \frac{2 \pi}{3}=120^{\circ}$ $\mathrm{LHS}=\cot x+\cot \left(60^{\circ}+x\right)+\cot \left(120^{\circ}+x\right)$ $=\cot x+\cot \left(60^{\circ}+x\right)-\cot \left[180^{\circ}-\left(120^{\circ}+x\right)\right]$ $\left(\because-\cot \theta=\cot \left(180^{\circ}-\theta\right)\right)$ $=\cot x+\cot \left(60^{\circ}+x\right)-\cot \left(60^{\circ}-x\right)$ $=\fra...
Read More →Find the missing frequency (p) for the following distribution whose mean is 7.68.
Question: Find the missing frequency (p) for the following distribution whose mean is 7.68. Solution: It is given that, Mean = 7.68 $\Rightarrow \sum f_{x} / N=7.68$ ⇒ 9p + 303p + 41 = 7.68 ⇒ 9p + 303 = 7.68p + 314.88 ⇒ 9p 7.68p = 314.88 303 ⇒ 1.32p = 11.88 ⇒ p = 11.881.32= 9 ⇒ p = 9 p = 9....
Read More →cot x+cot
Question: $\cot x+\cot \left(\frac{\pi}{3}+x\right)+\cot \left(\frac{\pi}{3}-x\right)=3 \cot 3 x$ Solution: $\frac{\pi}{3}=60^{\circ}$ $\mathrm{LHS}=\cot x+\cot \left(60^{\circ}+x\right)-\cot \left(60^{\circ}-x\right)$ $=\frac{1}{\tan x}+\frac{1}{\tan \left(60^{\circ}+x\right)}-\frac{1}{\tan \left(60^{\circ}-x\right)}$ $=\frac{1}{\tan x}+\frac{1-\sqrt{3} \tan x}{\sqrt{3}+\tan x}-\frac{1+\sqrt{3} \tan x}{\sqrt{3}-\tan x}$ $\left[\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}\right.$ and $\left....
Read More →Find the missing value of p for the following distribution whose mean is 12.58.
Question: Find the missing value of p for the following distribution whose mean is 12.58. Solution: It is given that, Mean = 12.58 $\Rightarrow \sum f_{X} / N=12.58$ $\Rightarrow \frac{7 p+524}{50}=12.58$ ⇒ 7p + 524 = 629 ⇒ 7p = 629 524 ⇒ 7p = 105 ⇒ p = 1057= 15 ⇒ p = 15 p = 18....
Read More →tan x
Question: $\tan x+\tan \left(\frac{\pi}{3}+x\right)-\tan \left(\frac{\pi}{3}-x\right)=3 \tan 3 x$ Solution: $\frac{\pi}{3}=60^{\circ}$ LHS $=\tan x+\tan \left(60^{\circ}+x\right)-\tan \left(60^{\circ}-x\right)$ $=\tan x+\left(\frac{\tan 60^{\circ}+\tan x}{1-\tan 60^{\circ} \tan x}\right)-\left(\frac{\tan 60^{\circ}-\tan x}{1+\tan 60^{\circ} \tan x}\right)$ $\left[\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}\right.$ and $\left.\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x \tan y}\right]$ $=\tan x+...
Read More →Find the vector equation of a plane which is at a distance of 7 units
Question: Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3 \hat{i}+5 \hat{j}-6 \hat{k}$. Solution: The normal vector is, $\vec{n}=3 \hat{i}+5 \hat{j}-6 \hat{k}$ $\therefore \hat{n}=\frac{\vec{n}}{|\vec{n}|}=\frac{3 \hat{i}+5 \hat{j}-6 \hat{k}}{\sqrt{(3)^{2}+(5)^{2}+(6)^{2}}}=\frac{3 \hat{i}+5 \hat{j}-6 \hat{k}}{\sqrt{70}}$ It is known that the equation of the plane with position vector $\vec{r}$ is given by, $\vec{r} \cdot \hat{n}=...
Read More →Prove that:
Question: Prove that: $\cos ^{3} x \sin 3 x+\sin ^{3} x \cos 3 x=\frac{3}{4} \sin 4 x$ Solution: We know, $\cos 3 x=4 \cos ^{3} x-3 \cos x$ $\Rightarrow \cos ^{3} x=\frac{\cos 3 x+3 \cos x}{4} \ldots$ (i) Also, $\sin 3 x=3 \sin x-4 \sin ^{3} x$ $\Rightarrow \sin ^{3} x=\frac{3 \sin x-\sin 3 x}{4} \ldots$ (ii) Now, LHS $=\cos ^{3} x \sin 3 x+\sin ^{3} x \cos 3 x$ $=\left(\frac{\cos 3 x+3 \cos x}{4}\right) \sin 3 x+\left(\frac{3 \sin x-\sin 3 x}{4}\right) \cos 3 x$ [Using (i) and (ii)] $=\frac{1}{...
Read More →If tan θ=2021, show that 1−sin θ+cos θ1+sin θ+cos θ=37.
Question: If $\tan \theta=\frac{20}{21}$, show that $\frac{1-\sin \theta+\cos \theta}{1+\sin \theta+\cos \theta}=\frac{3}{7}$. Solution: Given: $\tan \theta=\frac{20}{21}$.....(1) To show that: $\frac{1-\sin \theta+\cos \theta}{1+\sin \theta+\cos \theta}=\frac{3}{7}$ 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 Perpendic...
Read More →In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
Question: In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin. (a) $z=2$ (b) $x+y+z=1$ (c) $2 x+3 y-z=5$ (d) $5 y+8=0$ Solution: (a)The equation of the plane isz= 2 or 0x+ 0y+z= 2 (1) The direction ratios of normal are 0, 0, and 1. Dividing both sides of equation (1) by 1, we obtain $0 . x+0 . y+1 . z=2$ This is of the formlx+my+nz=d, wherel,m,nare the direction cosines of normal to the plane anddis the distance of the perpe...
Read More →Prove that:
Question: Prove that: $4\left(\cos ^{3} 10^{\circ}+\sin ^{3} 20^{\circ}\right)=3\left(\cos 10^{\circ}+\sin 20^{\circ}\right)$c Solution: We know, $\sin 60^{\circ}=\cos 30^{\circ} \quad\left(=\frac{\sqrt{3}}{2}\right)$ $\Rightarrow \sin 3 \times 20^{\circ}=\cos 3 \times 10^{\circ}$ $\Rightarrow 3 \sin 20^{\circ}-4 \sin ^{3} 20^{\circ}=4 \cos ^{3} 10^{\circ}-3 \cos 10^{\circ}$ $\left(\because \sin 3 \theta=3 \sin \theta-4 \sin ^{3} \theta\right.$ and $\left.\cos 3 \theta=4 \cos ^{3} \theta-3 \cos ...
Read More →Prove that:
Question: Prove that: $\sin 5 x=5 \sin x-20 \sin ^{3} x+16 \sin ^{5} x$ Solution: LHS $=\sin 5 x$ $=\sin (3 x+2 x)$ $=\sin 3 x \times \cos 2 x+\cos 3 x \times \sin 2 x$ $=\left(3 \sin x-4 \sin ^{3} x\right)\left(1-2 \sin ^{2} x\right)+\left(4 \cos ^{3} x-3 \cos x\right) \times 2 \sin x \cos x$ $=3 \sin x-6 \sin ^{3} x-4 \sin ^{3} x+8 \sin ^{5} x+\left(8 \cos ^{4} x-6 \cos ^{2} x\right) \sin x$ $=3 \sin x-10 \sin ^{3} x+8 \sin ^{5} x+\left\{8 \sin x\left(1-\sin ^{2} x\right)^{2}-6 \sin x\left(1-\...
Read More →If cosα+cosβ=0=sinα+sinβ,
Question: If $\cos \alpha+\cos \beta=0=\sin \alpha+\sin \beta$, then prove that $\cos 2 \alpha+\cos 2 \beta=-2 \cos (\alpha+\beta)$ [NCERT EXEMPLAR] Solution: Given: $\cos \alpha+\cos \beta=0=\sin \alpha+\sin \beta$ $\therefore \cos \alpha+\cos \beta=0$ Squaring on both sides, we get $\cos ^{2} \alpha+\cos ^{2} \beta+2 \cos \alpha \cos \beta=0$ ....(1) Also, $\sin \alpha+\sin \beta=0$ Squaring on both sides, we get $\sin ^{2} \alpha+\sin ^{2} \beta+2 \sin \alpha \sin \beta=0$ ...(2) Subtracting ...
Read More →If 3 cos θ − 4 sin θ = 2 cos θ + sin θ, find tan θ.
Question: If $3 \cos \theta-4 \sin \theta=2 \cos \theta+\sin \theta$, find $\tan \theta$. Solution: Given: $3 \cos \theta-4 \sin \theta=2 \cos \theta+\sin \theta$ To find: $\tan \theta$ Now consider the given expression $3 \cos \theta-4 \sin \theta=2 \cos \theta+\sin \theta$ Now by dividing both sides of the above expression by $\cos \theta$ We get, $\frac{3 \cos \theta-4 \sin \theta}{\cos \theta}=\frac{2 \cos \theta+\sin \theta}{\cos \theta}$ Now by separating the denominator for each terms We ...
Read More →If a cos2x+b sin2x=c has α and β as its roots,
Question: If $a \cos 2 x+b \sin 2 x=c$ has $\alpha$ and $\beta$ as its roots, then prove that (i) $\tan \alpha+\tan \beta=\frac{2 b}{a+c}$ [NCERT EXEMPLAR] (ii) $\tan \alpha \tan \beta=\frac{c-a}{c+a}$ (iii) $\tan (\alpha+\beta)=\frac{b}{a}$ [NCERT EXEMPLAR] Solution: Given: $a \cos 2 x+b \sin 2 x=c$ $\Rightarrow a\left(\frac{1-\tan ^{2} x}{1+\tan ^{2} x}\right)+b\left(\frac{2 \tan x}{1+\tan ^{2} x}\right)-c=0$ $\Rightarrow a\left(1-\tan ^{2} x\right)+2 b \tan x-c\left(1+\tan ^{2} x\right)=0$ $\...
Read More →If sin α=
Question: If $\sin \alpha=\frac{4}{5}$ and $\cos \beta=\frac{5}{13}$, prove that $\cos \frac{\alpha-\beta}{2}=\frac{8}{\sqrt{65}}$ Solution: Given: $\sin \alpha=\frac{4}{5}$ $\cos \beta=\frac{5}{13}$ Now, $\cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\sqrt{1-\left(\frac{4}{5}\right)^{2}}=\frac{3}{5}$ And, $\sin \beta=\sqrt{1-\cos ^{2} \alpha}=\sqrt{1-\left(\frac{5}{13}\right)^{2}}=\frac{12}{13}$ Now, $\cos (\alpha-\beta)=\cos \alpha \times \cos \beta+\sin \alpha \times \sin \beta$ $\Rightarrow \cos (\a...
Read More →If cos α + cos β=
Question: If $\cos \alpha+\cos \beta=\frac{1}{3}$ and $\sin \sin \alpha+\sin \beta=\frac{1}{4}$, prove that $\cos \frac{\alpha-\beta}{2}=\pm \frac{5}{24}$ Solution: Squaring and adding equations $\cos \alpha+\cos \beta=\frac{1}{3}$ and $\sin \alpha+\sin \beta=\frac{1}{4}$, we get $\Rightarrow 1+1+2(\cos \alpha \times \cos \beta+\sin \alpha \times \sin \beta)=\frac{25}{144}$ $\Rightarrow 2+2 \cos (\alpha-\beta)=\frac{25}{144} \quad(\because \cos (A-B)=\cos A \times \cos B+\sin A \times \sin B)$ $...
Read More →If sec (x+α)+sec (x−α)=2 sec x,
Question: If $\sec (x+\alpha)+\sec (x-\alpha)=2 \sec x$, prove that $\cos x=\pm \sqrt{2} \cos \frac{\alpha}{2}$ Solution: Equation sec $(x+\alpha)+\sec (x-\alpha)=2 \sec x$ can be written as $\frac{1}{\cos (x+\alpha)}+\frac{1}{\cos (x-\alpha)}=\frac{2}{\cos x}$ $\Rightarrow \frac{1}{\cos x \times \cos \alpha-\sin x \times \sin \alpha}+\frac{1}{\cos x \times \cos \alpha+\sin x \times \sin \alpha}=\frac{2}{\cos x}$ $[\because \cos (A+B)=\cos A \times \cos B-\sin A \times \sin B$ and $\cos (A-B)=\c...
Read More →Find the shortest distance between the lines whose vector equations are
Question: Find the shortest distance between the lines whose vector equations are $\vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k}$ and $\vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k}$ Solution: The given lines are $\vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k}$ $\Rightarrow \vec{r}=(\hat{i}-2 \hat{j}+3 \hat{k})+t(-\hat{i}+\hat{j}-2 \hat{k})$ ...(1) $\vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k}$ $\Rightarrow \vec{r}=(\hat{i}-\hat{j}-\hat{k})+s(\hat{i}+2 \hat{j}-2 \hat{k...
Read More →If cos x=
Question: If $\cos x=\frac{\cos \alpha+\cos \beta}{1+\cos \alpha \cos \beta}$, prove that $\tan \frac{x}{2}=\pm \tan \frac{\alpha}{2} \tan \frac{\beta}{2}$ Solution: Given: $\cos x=\frac{\cos \alpha+\cos \beta}{1+\cos \alpha \cos \beta}$ ....(1) $\left.\Rightarrow \frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}=\frac{\cos \alpha+\cos \beta}{1+\cos \alpha \times \cos \beta} \quad \because \cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right]$ By componendo and dividendo, w...
Read More →If 2 tan
Question: If $2 \tan \frac{\alpha}{2}=\tan \frac{\beta}{2}$, prove that $\cos \alpha=\frac{3+5 \cos \beta}{5+3 \cos \beta}$ Solution: $\mathrm{RHS}=\frac{3+5 \cos \beta}{5+3 \cos \beta}$ $=\frac{3+5\left(\frac{1-\tan ^{2} \frac{\beta}{2}}{1+\tan ^{2} \frac{\beta}{2}}\right)}{5+3\left(\frac{1-\tan ^{2} \frac{\beta}{2}}{1+\tan ^{2} \frac{\beta}{2}}\right)}$ $=\frac{3+3 \tan ^{2} \frac{\beta}{2}+5-5 \tan ^{2} \frac{\beta}{2}}{5+5 \tan ^{2} \frac{\beta}{2}+3-3 \tan \frac{\beta}{2}}$ $=\frac{8-2 \tan...
Read More →Find the value of p for the following distribution whose mean is 16.6.
Question: Find the value of p for the following distribution whose mean is 16.6. Solution: It is given that, Mean = 16.6 $\Rightarrow \frac{\sum \mathrm{fx}}{\mathrm{N}}=16.6$ $\Rightarrow \frac{24 \mathrm{p}+1228}{100}=16.6$ ⇒ 24p + 1228 = 1660 ⇒ 24p = 1660 1228 ⇒ 24p = 432 ⇒ p = 432/24= 18 ⇒ p = 18 p = 18....
Read More →If 8 tan A = 15, find sin A − cos A.
Question: If 8 tan A = 15, find sin A cos A. Solution: Given: $8 \tan A=15$ Therefore, $\tan A=\frac{15}{8}$....(1) To find: $\sin A-\cos A$ Now we know $\tan \theta$ is defined as follows $\tan A=\frac{\text { Perpendicular side opposite to } \angle A}{\text { Base side adjacent to } \angle A}$.....(4) Now by comparing equation (1) and (2) We get Perpendicular side opposite to $\angle A=15$ Base side adjacent to $\angle A=8$ Therefore triangle representing angle A is as shown below SideACis unk...
Read More →If the mean of the following data is 15, find p.
Question: If the mean of the following data is 15, find p. Solution: It is given that, Mean = 15 $\Rightarrow \sum f x / N=15$ ⇒10p + 445p + 27=15 ⇒10p + 445 = 15 (p + 27) ⇒ 10p + 445 = 15p + 405 ⇒15p 10p = 445 405 ⇒ 5p = 40 ⇒ p = 405= 8 ⇒ p = 8 p = 8....
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