Maths NCERT Exemplar Solutions Class 12th Chapter Two: Overview, Questions, Preparation

Sol. 

Q2. Find the simplified form of $co{s}^{-1}\left(\frac{3}{5}cos x+\frac{4}{5}sin x\right),�where x\in \left[\frac{-3\pi }{4},\frac{\pi }{4}\right].$

Sol.

$\begin{array}{l}4ta{n}^{-1}\frac{1}{5}-ta{n}^{-1}\frac{1}{239}\\ \Rightarrow 2.\left(2ta{n}^{-1}\frac{1}{5}\right)-ta{n}^{-1}\frac{1}{239}\\ \Rightarrow 2\left[ta{n}^{-1}\frac{2\times \frac{1}{5}}{1-\frac{1}{25}}\right]-ta{n}^{-1}\frac{1}{239}\left[2ta{n}^{-1}x=ta{n}^{-1}\frac{2x}{1-x^2}\right]\\ \Rightarrow 2ta{n}^{-1}\frac{5}{12}-ta{n}^{-1}\frac{1}{239}\Rightarrow ta{n}^{-1}\left(\frac{2\times \frac{5}{12}}{1-\frac{25}{144}}\right)-ta{n}^{-1}\frac{1}{239}\\ \Rightarrow ta{n}^{-1}\left(\frac{120}{119}\right)-ta{n}^{-1}\left(\frac{1}{239}\right)\\ \Rightarrow ta{n}^{-1}\left[\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119}\times \frac{1}{239}}\right]\left[\because ta{n}^{-1}x-ta{n}^{-1}y=ta{n}^{-1}\frac{x-y}{1+xy}\right]\\ \Rightarrow ta{n}^{-1}\left[\frac{120\times 239-119}{119\times 239+120}\right]=ta{n}^{-1}\left[\frac{28680-119}{28441+120}\right]\\ \Rightarrow ta{n}^{-1}\left[\frac{28561}{28561}\right]=ta{n}^{-1}\left(1\right)=\frac{\pi }{4}\end{array}$

Q8. If $a_1,a_2,a_3,\dots ,a_n$ is an arithmetic progression with common difference  $d$ , then evaluate the following expression:

tan[  $ta{n}^{-1}\left(\frac{d}{1+a_1{a}_2}\right)+ta{n}^{-1}\left(\frac{d}{1+a_2{a}_3}\right)+ta{n}^{-1}\left(\frac{d}{1+a_3{a}_4}\right)\cdots +ta{n}^{-1}\frac{d}{1+a_{n-1}{a}_n})].$

Q1. Find the value of $ta{n}^{-1}\left(tan\frac{5\pi }{6}\right)+co{s}^{-1}\left(cos\frac{13\pi }{6}\right).$

Sol:

$\begin{array}{l}Weknowthat\frac{5\pi }{6}\notin \left(-\frac{\pi }{2},\frac{\pi }{2}\right)and\frac{13\pi }{6}\notin \left[0,\pi \right]\\ \therefore ta{n}^{-1}\left(tan\frac{5\pi }{6}\right)+co{s}^{-1}\left(cos\frac{13\pi }{6}\right)\\ =ta{n}^{-1}\left[tan\left(\pi -\frac{\pi }{6}\right)\right]+co{s}^{-1}\left[cos\left(2\pi +\frac{\pi }{6}\right)\right]\\ =ta{n}^{-1}\left[tan\left(-\frac{\pi }{6}\right)\right]+co{s}^{-1}\left[cos\left(\frac{\pi }{6}\right)\right]\\ =ta{n}^{-1}\left(-tan\frac{\pi }{6}\right)+co{s}^{-1}\left(cos\frac{\pi }{6}\right)\\ =-ta{n}^{-1}\left(tan\frac{\pi }{6}\right)+co{s}^{-1}\left(cos\frac{\pi }{6}\right)\left[\because ta{n}^{-1}\left(-x\right)=-ta{n}^{-1}x\right]\\ =-\frac{\pi }{6}+\frac{\pi }{6}=0\\ Hence,ta{n}^{-1}\left(tan\frac{5\pi }{6}\right)+co{s}^{-1}\left(cos\frac{13\pi }{6}\right)=0\end{array}$

Sol.

Sol:

$\begin{array}{l}L.H.S.cot\left(\frac{\pi }{4}-2co{t}^{-1}3\right)\\ =cot\left[ta{n}^{-1}\left(1\right)-2ta{n}^{-1}\frac{1}{3}\right]\left[\because co{t}^{-1}x=ta{n}^{-1}\frac{1}{x}\right]\\ =cot\left[ta{n}^{-1}\left(1\right)-ta{n}^{-1}\frac{2\times \frac{1}{3}}{1-{\left(\frac{1}{3}\right)}^2}\right]\left[\because 2ta{n}^{-1}x=ta{n}^{-1}\frac{2x}{1-x^2}\right]\\ =cot\left[ta{n}^{-1}\left(1\right)-ta{n}^{-1}\frac{\frac{2}{3}}{\frac{8}{9}}\right]\\ =cot\left[ta{n}^{-1}\left(1\right)-ta{n}^{-1}\frac{3}{4}\right]\\ =cot\left[ta{n}^{-1}\left(\frac{1-\frac{3}{4}}{1+1\times \frac{3}{4}}\right)\right]=cot\left[ta{n}^{-1}\frac{\frac{1}{4}}{\frac{7}{4}}\right]\\ =cot\left[ta{n}^{-1}\frac{1}{7}\right]\left[\because ta{n}^{-1}\frac{1}{x}=co{t}^{-1}x\right]\\ =cot\left[co{t}^{-1}7\right]=7=R.H.S.\\ Henceproved.\end{array}$

Q4. Find the value of

Q5. Find the value of $ta{n}^{-1}\left(tan\frac{2\pi }{3}\right).$

Sol:

$\begin{array}{l}Weknowthat\frac{2\pi }{3}\notin \left[-\frac{\pi }{2},\frac{\pi }{2}\right]\\ \therefore ta{n}^{-1}\left(tan\frac{2\pi }{3}\right)=ta{n}^{-1}\left[tan\left(\pi -\frac{\pi }{3}\right)\right]=ta{n}^{-1}\left(-tan\frac{\pi }{3}\right)\\ =-ta{n}^{-1}\left(tan\frac{\pi }{3}\right)\left[\because ta{n}^{-1}\left(-x\right)=-ta{n}^{-1}x\right]\\ =-\frac{\pi }{3}\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]\\ Hence,ta{n}^{-1}\left(tan\frac{2\pi }{3}\right)=-\frac{\pi }{3}.\end{array}$

Sol:

$\begin{array}{l}L.H.S.2ta{n}^{-1}\left(-3\right)=-2ta{n}^{-1}\left(3\right)\\ =-co{s}^{-1}\left[\frac{1-{\left(3\right)}^2}{1+{\left(3\right)}^2}\right]\left[\because 2ta{n}^{-1}x=co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]\\ =-co{s}^{-1}\left[\frac{1-9}{1+9}\right]=-co{s}^{-1}\left(\frac{-8}{10}\right)\\ =-co{s}^{-1}\left(\frac{-4}{5}\right)=-\left[\pi -co{s}^{-1}\left(\frac{4}{5}\right)\right]=-\pi +co{s}^{-1}\left(\frac{4}{5}\right)\\ =-\pi +ta{n}^{-1}\left(\frac{3}{4}\right)\left[\because co{s}^{-1}\frac{4}{5}=ta{n}^{-1}\frac{3}{4}\right]\\ =-\pi +\frac{\pi }{2}-co{t}^{-1}\left(\frac{3}{4}\right)\left[\because ta{n}^{-1}x=\frac{\pi }{2}-co{t}^{-1}x\right]\\ =\frac{-\pi }{2}-co{t}^{-1}\left(\frac{3}{4}\right)\\ =\frac{-\pi }{2}-ta{n}^{-1}\left(\frac{4}{3}\right)\left[\because ta{n}^{-1}x=co{t}^{-1}\frac{1}{x}\right]\\ =\frac{-\pi }{2}+ta{n}^{-1}\left(-\frac{4}{3}\right)R.H.S.\\ Henceproved.\end{array}$

Q7. Find the real solutions of the equation

Sol:

Q8. Find the value of the expressions

Q9. If  $2 ta{n}^{-1}\left(cos \theta \right)=ta{n}^{-1}\left(2cosec \theta \right)$ , then show that  $\theta =\frac{\pi }{4}$ . where n is any integer.

Sol:

$\begin{array}{l}2ta{n}^{-1}\left(cos\theta \right)=ta{n}^{-1}\left(2cosec\theta \right)\\ \Rightarrow ta{n}^{-1}\left(\frac{2cos\theta }{1-co{s}^2\theta }\right)=ta{n}^{-1}\left(2cosec\theta \right)\\ \left[\because 2ta{n}^{-1}x=ta{n}^{-1}\frac{2x}{1-x^2}\right]\\ \Rightarrow \left(\frac{2cos\theta }{1-co{s}^2\theta }\right)=\left(2cosec\theta \right)\Rightarrow \frac{2cos\theta }{si{n}^2\theta }=\frac{2}{sin\theta }\\ \Rightarrow cos\theta sin\theta =si{n}^2\theta \\ \Rightarrow cos\theta sin\theta -si{n}^2\theta =0\Rightarrow sin\theta \left(cos\theta -sin\theta \right)=0\\ \Rightarrow sin\theta =0orcos\theta -sin\theta =0\\ \Rightarrow sin\theta =0or1-tan\theta =0\\ \Rightarrow \theta =0ortan\theta =1\\ \Rightarrow \theta =0^{0}or\theta =\frac{\pi }{4}Henceproved.\end{array}$

Q10. Show that $cos\left(2 ta{n}^{-1}\frac{1}{7}\right)=sin\left(4 ta{n}^{-1}\frac{1}{3}\right).$

Sol.

$\begin{array}{l}L.H.S.cos\left(2ta{n}^{-1}\frac{1}{7}\right)\\ =cos\left[co{s}^{-1}\frac{1-\frac{1}{49}}{1+\frac{1}{49}}\right]\left[\because 2ta{n}^{-1}x=co{s}^{-1}\frac{1-x^2}{1+x^2}\right]\\ =cos\left[co{s}^{-1}\frac{48}{50}\right]=cos\left[co{s}^{-1}\frac{24}{25}\right]=\frac{24}{25}\\ R.H.S.sin\left(4ta{n}^{-1}\frac{1}{3}\right)\\ =sin\left[2ta{n}^{-1}\left(\frac{2\times \frac{1}{3}}{1-\frac{1}{9}}\right)\right]\left[\because 2ta{n}^{-1}x=ta{n}^{-1}\frac{2x}{1-x^2}\right]\\ =sin\left[2ta{n}^{-1}\left(\frac{\frac{2}{3}}{\frac{8}{9}}\right)\right]=sin\left[2ta{n}^{-1}\frac{3}{4}\right]\\ =sin\left[si{n}^{-1}\left(\frac{2\times \frac{3}{4}}{1+\frac{9}{16}}\right)\right]\left[\because 2ta{n}^{-1}x=si{n}^{-1}\frac{2x}{1+x^2}\right]\\ =sin\left[si{n}^{-1}\frac{24}{25}\right]\Rightarrow \frac{24}{25}\\ L.H.S.=R.H.S.Henceproved.\end{array}$

Q11. Solve the following equation: $cos\left(ta{n}^{-1}x\right)=sin\left(co{t}^{-1}\frac{3}{4}\right).$