Maths

43. ${\displaystyle \sum _{}^{11}}\left(2+3^x\right)=\left(2+3^1\right)+\left(2+3^2\right)+.........+\left(2+3^{11}\right)$

$x=1$  [2+2+ ….+ (11lines)] + (31+ 32+……+311)

= $11*2+\frac{3\left(3^{11}-1\right)}{3-1}\left[?s_n=\frac{a\left(r^n-1\right)}{r-1},r\ge 1\right]$

= $22+\frac{3\left(3^{11}-1\right)}{2}$

41. Here,  $a_1=1$

$r=\frac{-9}{1}=-a$

So,  $s_n=\frac{a_1\left[1-r^n\right]}{1-r}$

= $\frac{1\left[1-{\left(-a\right)}^n\right]}{1-\left(-a\right)}$

= ${\frac{1-\left(a\right)}{1+a}}^n$

$ta{n}^{-1}\frac{x}{y}-ta{n}^{-1}\frac{x-y}{x+y}\left\{-ta{n}^{-1}x-ta{n}^{-1}y=\frac{x-y}{1+xy}\right\}$

$=ta{n}^{-1}\left\{\frac{\left(\frac{x}{y}\right)-\left(\frac{x-y}{x+y}\right)}{1+\frac{x}{y}·\left(\frac{x-y}{x+y}\right)}\right\}$

$=ta{n}^{-1}\left\{\frac{\frac{x(x+y)-(x-y)·y}{y(x+y)}}{\frac{y(x+y)+x(x-y)}{y(x+y)}}\right\}$

$=ta{n}^{-1}\left(\frac{x^2+xy-xy+y^2}{xy+y^2-x^2-xy}\right)$

$=ta{n}^{-1}\frac{x^2+y^2}{x^2+y^2}$

$=ta{n}^{-1}1$

$=ta{n}^{-1}\left(tan\frac{\pi }{4}\right)$

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

Option C is correct.

$si{n}^{-1}(1-x)-2si{n}^{-1}x=\frac{\pi }{2}\to (1)$

(M) Let $x=sin\theta .Then,\theta =si{n}^{-1}x.$

Putting this in $q^n(1)$ we get

$si{n}^{-1}(1-x)-2·\theta =\frac{\pi }{2}$

$\Rightarrow si{n}^{-1}(1-x)=\frac{\pi }{2}+2\theta$

$\Rightarrow 1-x=sin\left(\frac{\pi }{2}+2\theta \right)\left\{sin\left(\frac{\pi }{2}+x\right)=cosx\right\}$

$\Rightarrow 1-x=cos2\theta$

$\Rightarrow 1-x=1-2si{n}^2\theta ·\left\{cos2x=1-2si{n}^{2}x\right\}$

$\Rightarrow 1-x=1-2x^2·\{sin\theta =x\}$

$\begin{array}{l}\Rightarrow 2x^2-x=0\\ \Rightarrow x(2x-1)=0\end{array}$

$\Rightarrow so,x=0x2x-1=0x2x=1\to x=\frac{1}{2}.$

Putting $x=0$ in qⁿ (1) .

L.H.S $=si{n}^{-1}(1-0)-2si{n}^{-1}0=si{n}^{-1}sin\frac{x}{2}-0=\frac{\pi }{2}=R.H.S.$

$x=\frac{1}{2}\mathrm{q(1)}$

$L.H.S=si{n}^{-1}\left(1-\frac{1}{2}\right)-2si{n}^{-1}\frac{1}{2}=si{n}^{-1}\left(\frac{1}{2}\right)-2si{n}^{-1}\frac{1}{2}$

$=si{n}^{-1}\frac{1}{2}-2si{n}^{-1}\frac{1}{2}$

$=-si{n}^{-1}\frac{1}{2}=-si{n}^{-1}\left(sin\frac{\pi }{6}\right)=-\frac{\pi }{6}$ $\ne$

So, $=0.$

Option (c) is correct.

Given, (M)

$ta{n}^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}ta{n}^{-1}x$

$x=tan\mathrm{\theta .Then}\theta =ta{n}^{-1}\mathrm{x\; Bo\; we\; have,}$

$ta{n}^{-1}\left(\frac{1-tan\theta }{1+tan\theta }\right)=\frac{1}{2}ta{n}^{-1}(tan\theta )$

$\Rightarrow ta{n}^{-1}\left(\frac{tan\frac{\pi }{4}-tan\theta }{1+tan\frac{x}{4}tan\theta }\right)=\frac{1}{2}\theta \left\{?tan\frac{\pi }{4}=1\right\}.$

$\Rightarrow ta{n}^{-1}\left\{tan\left(\frac{x}{4}-\theta \right)\right\}=\frac{\theta }{2}\{\frac{?tanx-tany}{1+tanxtan}=tan(x-y)$

$\Rightarrow \frac{\pi }{4}-\theta =\frac{\theta }{2}$

$\Rightarrow \frac{\theta }{2}+\theta =\frac{x}{4}$

$\Rightarrow \frac{3\theta }{2}=\frac{\pi }{4}$

$\Rightarrow \theta =\frac{\pi }{4}*\frac{2}{3}=\frac{\pi }{6}$

Given,

$(M)2ta{n}^{-1}(cosx)=ta{n}^{-1}(2cosecx)$

$\Rightarrow ta{n}^{-1}\frac{2cosx}{1-co{s}^{2}x}=ta{n}^{-1}\frac{2}{sinx}$ $\left\{\begin{array}{l}using.\\ ta{n}^{-1}2x=\frac{2x}{1-x^2}\end{array}\right\}$

$\Rightarrow \frac{2cosx}{si{n}^{2}x}=\frac{2}{sinx}$ $\left\{?1-co{s}^{2}x=si{n}^{2}x\Rightarrow 1=si{n}^{2}x+co{s}^{2}x\right\}$

$\Rightarrow \frac{cosx}{sinx}=1$

$\Rightarrow cotx=cot\frac{x}{4}$

$\Rightarrow x=\frac{\pi }{4}.$

39. Here a = 0.15 = $\frac{15}{100}$

$r=\frac{0.015}{0.15}=\frac{\frac{15}{1000}}{\frac{15}{150}}=\frac{1}{10}$ <1

n = 20

So, Sum to term of G.P., $s_n=\frac{a(1-r^n)}{1-r}$

$\Rightarrow$ $s_{20}=\frac{\frac{15}{100}\left[1-{\left(\frac{1}{10}\right)}^{20}\right]}{1-\frac{1}{10}}$

= $\frac{\frac{15}{100}\left[1-{(0.1)}^{20}\right]}{\frac{9}{10}}$

= $\frac{15}{100}*\frac{10}{9}\left[1-{(0.1)}^{20}\right]$

= $\frac{1}{6}\left[1-{(0.1)}^{20}\right]$

LH.S $=\left(ta{n}^{-1}\frac{1}{5}+ta{n}^{-1}\frac{1}{7}\right)+\left(ta{n}^{-1}\frac{1}{3}+ta{n}^{-1}\frac{1}{8}\right)$

$=ta{n}^{-1}\left[\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5}·\frac{1}{7}}\right]+ta{n}^{-1}\left[\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3},\frac{1}{8}}\right]\left\{\begin{array}{cc}\hfill ?usingta{n}^{-1}x+ta{n}^{-1}y\hfill & \hfill \frac{x+y}{1-xy},xy<1\hfill \end{array}\right\}$

= $ta{n}^{-1}\left[\frac{\frac{7+5}{7*5}}{\frac{7*5-1}{7*5}}\right]+ta{n}^{-1}\left[\frac{\frac{8+3}{5*3}}{\frac{8*3-1}{8*3}}\right]$

$=ta{n}^{-1}\left(\frac{12}{35-1}\right)+ta{n}^{-1}\left(\frac{11}{24-1}\right)=ta{n}^{-1}\frac{12}{34}+ta{n}^{-1}\frac{11}{23}$

$=ta{n}^{-1}\frac{6}{17}+ta{n}^{-1}\frac{11}{23}$

$=ta{n}^{-1}\left(\frac{\frac{6}{17}+\frac{1}{23}}{1-\frac{6}{17}*\frac{3}{23}}\right)=ta{n}^{-1}\left(\frac{\frac{6*23+11*11}{17*23}}{\frac{?7*23-6*11}{17*23}}\right)$

$=ta{n}^{-1}\frac{138+187}{391-66}=ta{n}^{-1}\frac{325}{325}=ta{n}^{-1}1$

$=ta{n}^{-1}\left(tan\frac{1}{4}\right)$

$=\frac{x}{4}=R.H.S$

Let $si{n}^{-1}\frac{5}{13}=x\text{\hspace{0.17em}and}co{s}^{-1}\frac{3}{5}=y.$

Then, $sinx=\frac{5}{13}andcos=\frac{3}{5}$

So, $tanx=\frac{sinx}{cosx}andtany=\frac{siny}{cosy}$

$=\frac{5/3}{12/1}=\frac{4/5}{3/5}.$

$=\frac{5}{12}=\frac{4}{3}.$

Using $tan(x+y)=\frac{lanx+lany}{1-tanxtany.}$

tan $\left(si{n}^{-1}\frac{5}{13}+co{s}^{-1}\frac{3}{5}\right)=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}*\frac{4}{3}}=$ $\frac{\frac{5*3+4*12}{12*3}}{\frac{12*3-5*4}{12*3}}$

$=\frac{15+48}{36-20}=\frac{63}{16}$

$\Rightarrow si{n}^{-1}\frac{5}{13}+co{s}^{-1}\frac{3}{5}=ta{n}^{-1}\frac{63}{16}.$

Hence proved.