Maths

34.$=\frac{sin5x+sin3x}{cos5x+cos3x.}$

$=\frac{2sin\left(\frac{5x+3x}{2}\right)cos\left(\frac{5x-3x}{2}\right)}{2cos\left(\frac{5x+3x}{2}\right)cos\left(\frac{5x-3x}{2}\right)}$

$=\frac{sin\frac{8x}{2}cos\frac{2x}{2}}{cos\frac{8x}{2}cos\frac{2x}{2}}$

$=\frac{sin4x}{cos4x}=tan4x=$R.H.S

The coordinates of the points, O and P, are (0, 0) and (1, 2, −3) respectively.

Therefore, the direction ratios of OP are (1 − 0) = 1, (2 − 0) = 2, and (−3 − 0) = −3

It is known that the equation of the plane passing through the point (x1,  y1 z1) is

$a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0$ where, a,  b, and c are the direction ratios of normal.

Here, the direction ratios of normal are 1, 2, and −3 and the point P is (1, 2, −3).

Thus, the equation of the required plane is

$\begin{array}{l}\mathrm{}\hfill \end{array}$$\begin{array}{ll}1\left(x-1\right)+2\left(y-2\right)-3\left(z+3\right)0\hfill & \Rightarrow x+2y-3z-14=0\hfill \end{array}$

33.$=\frac{cos9x-cos5x}{sin17x-sin3x}$

$=\frac{-2sin\left(\frac{9x+5x}{2}\right)sin\left(\frac{9x-5x}{2}\right)}{2cos\left(\frac{17x+3x}{2}\right)sin\left(\frac{17x-3x}{2}\right)}$

$=\frac{-sin\frac{14x}{2}sin\frac{4x}{2}}{cos\frac{20x}{2}sin\frac{14x}{2}}=\frac{-sin7xsin2x}{cos10xsin7x}=\frac{-sin2x}{cos10x}$

= R.H.S

Equation of one plane is

$\begin{array}{l}\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=1\\ \Rightarrow \overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)-1=0\\ \overrightarrow{r}.\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)+4=0\end{array}$

The equation of any plane passing through the line of intersection of these planes is

$\begin{array}{l}\left[\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)-1\right]+\lambda \left[\overrightarrow{r}.\left(2\widehat{i}+3\widehat{j}-\widehat{k}\right)+4\right]=0\\ \overrightarrow{r}.\left[\left(2\lambda +1\right)\widehat{i}+\left(3\lambda +1\right)\widehat{j}+\left(1-\lambda \right)\widehat{k}\right]+\left(4\lambda -1\right)=0.....(1)\end{array}$

Its direction ratios are (2λ + 1), (3λ + 1), and (1 − λ).

The required plane is parallel to x-axis. Therefore, its normal is perpendicular to x-axis.

The direction ratios of x-axis are 1, 0, and 0.

$\begin{array}{l}\therefore 1.\left(2\lambda +1\right)+0\left(3\lambda +1\right)+0\left(1-\lambda \right)=0\\ \Rightarrow 2\lambda +1=0\\ \Rightarrow \lambda =-\frac{1}{2}\end{array}$

Substituting λ = -1/2 in equation (1), we obtain

$\begin{array}{l}\Rightarrow \overrightarrow{r}.\left[-\frac{1}{2}\widehat{j}+\frac{3}{2}\widehat{k}\right]+\left(-3\right)=0\\ \Rightarrow \overrightarrow{r}\left(\widehat{j}-3\widehat{k}\right)+6=0\end{array}$

Therefore, its Cartesian equation is y − 3z + 6 = 0

This is the equation of the required plane.

The position vector through the point $\left(1,1, p\right)$ is  $\overrightarrow{a_1}=\widehat{i}+\widehat{j}+p\widehat{k}$

Similarly, the position vector through the point $\left(-3,0,1\right)$ is

$\overrightarrow{a_2}=-4\widehat{i}+\widehat{k}$

The equation of the given plane is  $\overrightarrow{r}.\left(3\widehat{i}+4\widehat{j}-12\widehat{k}\right)+13=0$

It is known that the perpendicular distance between a point whose position vector is  $\overrightarrow{a}$  and the plane,  $\overrightarrow{r}.\overrightarrow{N}=d$ is given by,  $D=\frac{\left|\overrightarrow{a}.\overrightarrow{N}-d\right|}{\left|\overrightarrow{N}\right|}$

Here, $\overrightarrow{N}=3\widehat{i}+4\widehat{j}-12\widehat{k}$ and  $d =-13$

Therefore, the distance between the point (1, 1, p) and the given plane is

$\begin{array}{l}{D}_1=\frac{\left|\left(\widehat{i}+\widehat{j}+p\widehat{k}\right).\left(3\widehat{i}+4\widehat{j}-12\widehat{k}\right)+13\right|}{\left|3\widehat{i}+4\widehat{j}-12\widehat{k}\right|}\\ \Rightarrow D_1=\frac{\left|3+4-12p+13\right|}{}\\ \Rightarrow D_1=\frac{\left|20-12p\right|}{13}..........(1)\end{array}$

Similarly, the distance between the point $\left(-3,0,1\right)$ and the given plane is

$\begin{array}{l}{D}_2=\frac{\left|\left(3\widehat{i}+\widehat{k}\right).\left(3\widehat{i}+4\widehat{j}-12\widehat{k}\right)+13\right|}{\left|3\widehat{i}+4\widehat{j}-12\widehat{k}\right|}\\ \Rightarrow D_1=\frac{\left|-9-12+13\right|}{}\\ \Rightarrow D_1=\frac{8}{13}..........(2)\end{array}$

It is given that the distance between the required plane and the points, $\left(1,1, p\right)and\left(-3,0,1\right),$ is equal.

$\therefore D_1 = D_2$

$\begin{array}{l}\Rightarrow \frac{\left|20-12p\right|}{13}=\frac{8}{13}\\ \Rightarrow 20-12p=8,or,-(-20-12p)=8\\ \Rightarrow 12p=12,or,12p=28\\ \Rightarrow p=1,or,p=\frac{7}{3}\end{array}$

32. L.H.S$=cot4x(sin5x+sin3x)$

$=cot4x\left[2sin\frac{8x}{2}cos\frac{2x}{2}\right]$

$=\frac{cos4x}{4x}*2sin4xcosx$

$=2·cos4x·cosx$

R.H.S$=cotx[sin5x-sin3x]$

$=cotx\left[2cos\frac{8x}{2}·sin\frac{2x}{2}\right]$

$=\frac{cosx}{sinx}·2cos4xsinx$

$=2·cos4x·cosx.$

Hence, L.H.S. = R.H.S.

The equation of the plane passing through the point $a \left(x +1\right)+ b \left(y -3\right)+ c \left(z -2\right)=0\dots \left(1\right)$ where, a, b, c are the direction ratios of normal to the plane.

It is known that two planes,  $a_{1}x+b_{1}y+c_{1}z+d_1=0\&a_{2}x+b_{2}y+c_{2}z+d_2=0$ are perpendicular, if $a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

Plane (1) is perpendicular to the plane,  $x +2y +3z =5$

$\begin{array}{ll}a.1 + b .2 + c.3 = 0\hfill & a + 2b + 3c = 0 ....\left(2\right)\hfill \end{array}$

Also, plane (1) is perpendicular to the plane, $3x +3y + z =0$

$\begin{array}{ll}a .3 + b.3 + c.1 = 0\hfill & 3a + 3b + c = 0 .....\left(3\right)\hfill \end{array}$

From equations (2) and (3), we obtain

$\begin{array}{l}\frac{a}{2*1-3*3}=\frac{b}{3*3-1*1}=\frac{c}{1*3-2*3}\\ \Rightarrow \frac{a}{-7}=\frac{b}{8}=\frac{c}{-3}=k(say)\\ \Rightarrow a=-7k,b=8k,c=-3k\end{array}$

Substituting the values of a, b, and c in equation (1), we obtain

$\begin{array}{l}-7k(x+1)+8k(y-3)-3k(z-2)=0\\ \Rightarrow (-7x-7)+(8y-24)-3z+6=0\\ \Rightarrow -7x+8y-3z-25=0\\ \Rightarrow 7x-8y+3z+25=0\end{array}$

This is the required equation of the plane.

It is known that the equation of the line through the points, $(x_1, y_1, z_1)and(x_2, y_2, z_2)$ , is

$\frac{x-x_1}{x_2-x_1}=\frac{x-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

Since the line passes through the points, $\left(3,-4,-5\right)and\left(2,-3,1\right)$ , its equation is given by,

$\begin{array}{l}\frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}\\ \Rightarrow \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=k(say)\\ \Rightarrow x=3-k,y=k-4,z=6k-5\end{array}$

Therefore, any point on the line is of the form $\left(3- k, k -4,6k -5\right).$

This point lies on the plane, $2x + y + z =7$

$\begin{array}{lll}\therefore 2 \left(3- k\right) + \left(k -4\right) + \left(6k -5\right) = 7\hfill & 5k - 3 = 7\hfill & k = 2\hfill \end{array}$

Hence, the coordinates of the required point are $\left(3-2,2-4,6*2-5\right)i.e.,$ $\left(1,-2,7\right).$

31. L.H.S. = sin 2x + 2 sin 4x + sin 6x

Using sin A + sin B = 2 sin A + B/2 cos A - B/2  we have,

L.H.S. = (sin 2x + sin 6x) + 2 sin 4x

$=2sin\left(\frac{2x+6x}{2}\right)cos\left(\frac{2x-6x}{2}\right)+2sin4x.$

$=2sin\frac{8x}{2}cos\left(-\frac{4x}{2}\right)+2sin4x.$

$=2sin4xcos2x+2sin4x[?cos(-x)=x]$

$=2sin4x[cos2x+1]$

We know that,

$cos2x=2co{s}^{2}x-1$

$\Rightarrow cos2x+1=2co{s}^{2}x$

Hence,

L.H.S$=2sin4x\left(2co{s}^{2}x\right)$

$=4co{s}^{2}xsin4x$

= R.H.S

It is known that the equation of the line passing through the points, $(x_1, y_1, z_1)and(x_2, y_2, z_2),$ is  $\frac{x-x_1}{x_2-x_1}=\frac{x-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

The line passing through the points, $\left(5,1,6\right)and\left(3,4,1\right),$ is given by,

$\begin{array}{l}\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6}\\ \Rightarrow \frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=k(say)\\ \Rightarrow x=5-2k,y=3k+1,z=6-5k\end{array}$

Any point on the line is of the form $\left(5-2k,3k +1,6-5k\right).$

Since the line passes through ZX-plane,

$\begin{array}{l}3k+1=0\\ \Rightarrow k=-\frac{1}{3}\\ \Rightarrow 5-2k=5-2\left(-\frac{1}{3}\right)=\frac{17}{3}\\ 6-5k=6-5\left(-\frac{1}{3}\right)=\frac{23}{3}\end{array}$

Therefore, the required point is $\left(\frac{17}{3},0,\frac{23}{3}\right)$