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RD Sharma Class 12 Exercise 3.11 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 3.11 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 3.11 Inverse Trigonometric Functions Solutions Maths - Download PDF Free Online

Updated on 21 Jan 2022, 11:34 AM IST

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RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise

Inverse Trigonometric Functions Excercise: 3.11

Inverse Trigonometric Function Exercise 3.11 Question 1(i)

Answer:
$\tan ^{-1}\left ( \frac{2}{9} \right )$
Hint:
There is one formula in trigonometric function for union.
Let’s see the formula,
$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left ( \frac{x+y}{1-xy} \right )$
Given:
$\tan^{-1}\left ( \frac{1}{7} \right )+\tan^{-1}\left ( \frac{1}{13} \right )=\tan^{-1}\left ( \frac{2}{9} \right )$
In LHS side,
Let $x=\frac{1}{7} \: and\: y=\frac{1}{13}$
Explanation:
Let’s take LHS
L.H.S $=\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{13}$
Let’s use formula,
$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left ( \frac{x+y}{1-xy} \right )$
Let’s put the value of x and y in formula
$\begin{aligned} L.H.S &=\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13} \\ &=\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{1}{13}}{1-\left(\frac{1}{7} \times \frac{1}{13}\right)}\right) \end{aligned}$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{13+7}{91}}{\frac{91-1}{91}}\right) \\ &=\tan ^{-1}\left(\frac{13+7}{91-1}\right) \\ &=\tan ^{-1}\left(\frac{20}{90}\right) \\ &=\tan ^{-1}\left(\frac{2}{9}\right) \end{aligned}$
$=R.H.S$
Hence, the prove
Note: We must remember the formula of union.

Inverse Trigonometric Function Exercise 3.11 Question 1(ii).

Answer:
$\pi$
Hint:
To solve this type of question, we must convert $\sin$ and $\cos$ in $\tan$ using hypotenuse theorem.
Given:
$\sin^{-1}\left ( \frac{12}{13} \right )+\cos^{-1}\left ( \frac{4}{5} \right )+\tan^{-1}\left ( \frac{63}{16} \right )=\pi$
Explanation:
For proving this type of question,
Let’s assume,
$\! \! \! \! \! \! \! \! \sin^{-1}\left ( \frac{12}{13} \right )= \alpha \\\cos^{-1}\left ( \frac{4}{5} \right )= \beta \\\tan^{-1}\left ( \frac{63}{16} \right )=\gamma$
So, we have to prove that
$\alpha+ \beta+ \gamma = \pi$
First of all
$\sin\alpha =\frac{12}{13}$

Using the formula of hypotenuse , third side will be 5.
$\tan\alpha =\frac{12}{5} \cdot \cdot \cdot (i)$
Also we have,
$\cos\beta =\frac{4}{5}$

Using the formula of hypotenuse, third side will be 3
$\tan\beta =\frac{3}{4}$ …(ii)
Also we have,
$\! \! \! \! \! \! \! \! \tan \gamma =\frac{63}{16} \cdot \cdot \cdot \cdot (iii)\\ \alpha +\beta +\gamma =\pi \\ \alpha +\beta =\pi -\gamma \\$
Let’s take tan on both sides,
$\tan\alpha +\beta =\tan\pi -\gamma$
$\frac{\tan \alpha +\tan \beta}{1-\ tan\alpha \tan \beta } =\frac{\tan\pi +\tan\gamma }{ 1- \tan\pi \tan\gamma }$
Let’s put value, from (i), (ii) and (iii)
$\begin{aligned} &\Rightarrow \frac{\frac{12}{5}+\frac{3}{4}}{1-\left(\frac{12}{5} \times \frac{3}{4}\right)}=\frac{\tan \pi-\tan \gamma}{1-\tan \gamma \tan \pi} \\ &\Rightarrow \frac{\frac{12}{5}+\frac{3}{4}}{1-\left(\frac{12}{5} \times \frac{3}{4}\right)}=\frac{-63}{16} \\ &\Rightarrow \frac{48+15}{-16}=\frac{-63}{16} \\ &\Rightarrow \frac{-63}{16}=\frac{-63}{16} \end{aligned}$
∴ L.H.S =R.H.S
Hence, the prove
Note: This type of problem can be easier by $\alpha+ \beta+ \gamma = \pi$

Inverse Trigonometric Function Exercise 3.11 Question 1(iii).

Answer:
$\sin^{-1}\left ( \frac{1}{\sqrt{5}} \right )$
Hint:
For solving this, we can use the formula of union trigonometric function,
$\tan^{-1}a+\tan^{-1}b=\tan^{-1}\left ( \frac{a+b}{1-ab} \right )$
Given:
$\tan^{-1}\left ( \frac{1}{4} \right )+\tan^{-1}\left ( \frac{2}{9} \right )=\sin^{-1}\left ( \frac{1}{\sqrt{5}} \right )$
Explanation:
L.H.S
$= \tan^{-1}\left ( \frac{1}{4} \right )+\tan^{-1}\left ( \frac{2}{9} \right )$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\left(\frac{1}{4} \times \frac{2}{9}\right)}\right) \\ &=\tan ^{-1}\left(\frac{\frac{9+8}{36}}{\frac{36-2}{36}}\right) \\ &=\tan ^{-1}\left(\frac{17}{34}\right) \end{aligned}$
$=\tan^{-1}\left ( \frac{1}{2} \right )$
Let’s take $\tan^{-1}\left ( \frac{1}{2} \right )= \theta$
So,$\tan\left ( \frac{1}{2} \right )= \theta$ [Removing inverse]
$\tan\theta =\left ( \frac{1}{2} \right ) = \frac{Perp}{Base} . \cdot \cdot \cdot \cdot (i)$

Using the formula of hypotenuse , Hypotenuse =$\sqrt{5}$
$\! \! \! \! \! \! \! \! \sin\theta =\frac{P}{H}=\frac{1}{\sqrt{5}}\\ \theta =\sin^{-1}\left ( \frac{1}{\sqrt{5}} \right ) \cdot \cdot \cdot \cdot (ii)$
Let’s put value of (ii) into (i)
$\! \! \! \! \! \! \! \! \! \tan\left \{ \sin^{-1}\left ( \frac{1}{\sqrt{5}} \right ) \right \}=\frac{1}{2}\\ \tan^{-1}\left ( \frac{1}{2} \right )=\sin^{-1}\left ( \frac{1}{\sqrt{5}} \right )$ … (iii)
So, $\tan^{-1}\left ( \frac{1}{2} \right )=\sin^{-1}\left ( \frac{1}{\sqrt{5}} \right )$ =R.H.S
Hence, the prove
Note: We must remember the formula of hypotenuse.

Inverse Trigonometric Function Exercise 3.11 Question 2.

Answer:
$\frac{\pi }{4}$
Hint:
For solving this, we can use the formula of union trigonometric function,
$\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left ( \frac{a-b}{1+ab} \right )$
Given:
$\tan^{-1}\left ( \frac{x}{y} \right )-\tan^{-1}\left ( \frac{x-y}{x+y} \right )$
Explanation:
Let’s use the formula
Where$a=\frac{x}{y}\: and \: b=\frac{x-y}{x+y}$
$\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right)=\tan ^{-1}\left(\frac{\frac{x}{y}-\left(\frac{x-y}{x+y}\right)}{1+\left(\frac{x}{y}\right)\left(\frac{x-y}{x+y}\right)}\right)$
$\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{x^{2}+x y-y(x-y)}{y(x+y)}}{\frac{x y+y^{2}+x^{2}-x y}{y(x+y)}}\right) \\ &=\tan ^{-1}\left(\frac{x^{2}+y^{2}}{x^{2}+y^{2}}\right) \\ &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}$
Note: We must know the formula of union trigonometric function.

Inverse Trigonometric Function Exercise 3.11 Question 3(i).

Answer:
$x=\left ( \frac{-1}{6} \right ) \: or \: x= 1$
For solving this, we can use the formula of union trigonometric function,
$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB } \right )$
Given:
$\tan^{-1}2x+\tan^{-1}3x=nx+\frac{3\pi }{4}$ and we have to find the value of x.
Solution:
Here A=2x and B=3x
So, let’s put the values of A and B in the formula of$\tan^{-1}A+ \tan^{-1}B$
$\begin{aligned} &\Rightarrow \tan ^{-1} 2 x+\tan ^{-1} 3 x=n x+\frac{3 \pi}{4} \\ &\Rightarrow \tan ^{-1}\left(\frac{2 x+3 x}{1-(2 x)(3 x)}\right)=n x+\frac{3 \pi}{4} \end{aligned}$
$\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{5 x}{1-6 x^{2}}\right)=n \pi+\frac{3 \pi}{4} \\ &\Rightarrow \frac{5 x}{1-6 x^{2}}=\tan \left(n \pi+\frac{3 \pi}{4}\right) \\ &\Rightarrow \frac{5 x}{1-6 x^{2}}=\frac{\tan (n \pi)+\tan \frac{3 \pi}{4}}{1-\tan (n \pi) \tan \frac{3 \pi}{4}} \end{aligned} \quad\left [\because \operatorname{tann} \pi=\mathbf{0}, \tan \frac{3\pi }{4}= -1 \right ]$
$\begin{aligned} &\Rightarrow \frac{5 x}{1-6 x^{2}}=-1 \\ &\Rightarrow 5 x=6 x^{2}-1 \\ &\Rightarrow 6 x^{2}-6 x+x-1=0 \\ &\Rightarrow 6 x(x-1)+1(x-1)=0 \\ &\Rightarrow(6 x+1)(x-1)=0 \end{aligned}$
$\Rightarrow Either\: 6x+1=0 \: or \: x -1=0$
$\Rightarrow 6x=-1 \: or \: x=1$
$\Rightarrow x=\frac{-1}{6} , 1$

Inverse Trigonometric Function Exercise 3.11 Question 3(ii).

Answer:
$x=\frac{1}{4},-8$
Hint:
Here, we can use the below formula,
$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB} \right )$
Given:
$\tan^{-1}\left ( x+1 \right )+\tan^{-1}\left (x-1 \right )=\tan^{-1}\left (\frac{8}{31} \right )$
Solution:
Here, A=x+1
B=x-1
So, let’s put the values of A and B in the formula of $\tan^{-1}A+\tan^{-1}B$
$\! \! \! \! \! \! \! \! \! \Rightarrow \tan^{-1}\left (x+1 \right )+\tan^{-1}\left ( x-1\right ) = \tan^{-1}\left ( \frac{8}{31} \right )\\ \Rightarrow \tan^{-1}\left ( \frac{x+1+x-1}{1-\left (x+1 \right )\left ( x-1 \right )} \right )=\tan^{-1}\frac{8}{31}\\ \Rightarrow \tan^{-1}\left ( \frac{2x}{1-\left ( x^{2}-1 \right )} \right )=\tan^{-1}\frac{8}{31}\\ \Rightarrow \frac{2x}{1-x^{2}+1}=\frac{8}{31}\\ \Rightarrow 31x=8-4x^{2}$
$\! \! \! \! \! \! \! \! \! \Rightarrow 4x^{2}+31x-8=0\\ \Rightarrow 4x^{2}+32x-x-8=0\\ \Rightarrow 4x\left ( x+8 \right )-1\left ( x+8 \right )=0\\ \Rightarrow 4x-1=0 \: or \: x+8=0\\ \Rightarrow x=\frac{1}{4} \: or \: x=- 8\\ \Rightarrow x=\frac{1}{4},-8$

Inverse Trigonometric Function Exercise 3.11 Question 3(iii).

Answer:
$x=\left ( \frac{-1}{2} \right ),\left ( \frac{1}{2} \right )$
Hint:
Here, we can use the formula of union trigonometric function,
$\tan^{-1}a+\tan^{-1}b=\tan^{-1}\left ( \frac{a+b}{1-ab} \right )$
Given:
We have to solve$\tan^{-1}\left ( x-1 \right )+ \tan^{-1}x+ \tan^{-1}\left ( x+1 \right )= \tan^{-1}3x$ and find the value of x.
Solution:
L.H.S=R.H.S
$\tan^{-1}\left ( x-1 \right )+ \tan^{-1}x+ \tan^{-1}\left ( x+1 \right )= \tan^{-1}3x$
Let’s use in sum formula,
$\begin{aligned} &\tan ^{-1}\left(\frac{x-1+x+1}{1-\left(x^{2}-1\right)}\right)+\tan ^{-1} x=\tan ^{-1} 3 x \\ &\Rightarrow \tan ^{-1}\left(\frac{2 x}{2-x^{2}}\right)=\tan ^{-1} 3 x-\tan ^{-1} x \\ &\Rightarrow \tan ^{-1}\left(\frac{2 x}{2-x^{2}}\right)=\tan ^{-1}\left(\frac{3 x-x}{1+3 x^{2}}\right) \\ &\Rightarrow \frac{2 x}{2-x^{2}}=\frac{3 x-x}{1+3 x^{2}} \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{2 x}{2-x^{2}}=\frac{2 x}{1+3 x^{2}} \\ &\Rightarrow 1+3 x^{2}=2-x^{2} \\ &\Rightarrow 3 x^{2}+x^{2}=2-1 \\ &\Rightarrow 4 x^{2}=1 \\ &\Rightarrow x^{2}=\frac{1}{4} \\ &\Rightarrow x=0, \pm \frac{1}{2} \end{aligned}$
$x=\left ( \frac{-1}{2} \right ),\left ( \frac{1}{2} \right ), 0$

Note: We must know the formula of Intersection.
$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB} \right )$

Inverse Trigonometric Function Exercise 3.11 Question 3(iv).

Answer:
$x = \frac{1}{\sqrt{3}}$
Hint:
Here, we use the formula of Intersection
$\tan^{-1}A-\tan^{-1}B=\tan^{-1}\left ( \frac{A-B}{1+AB} \right )$
Given:
$\tan^{-1}\left ( \frac{1-x}{1+x} \right )-\frac{1}{2}\tan^{-1}x=0$
Solution:
Here let’s assume
$\begin{aligned} &x=\tan \theta \\ &\Rightarrow \tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)-\frac{1}{2} \tan ^{-1}(\tan \theta)=0 \\ &\Rightarrow \tan ^{-1}\left(\frac{\tan \frac{\pi}{4}-\tan \theta}{1+\left(\tan \frac{\pi}{4}\right)(\tan \theta)}\right)-\frac{1}{2} \theta=0 \\ &\Rightarrow \tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\theta\right)\right]-\frac{\theta}{2}=0 \end{aligned}$
$\begin{aligned} &\Rightarrow \frac{\pi}{4}-\theta-\frac{\theta}{2}=0 \\ &\Rightarrow \frac{3 \theta}{2}=\frac{\pi}{4} \\ &\Rightarrow \theta=\frac{\pi}{6} \\ &\Rightarrow \tan ^{-1} x=\frac{\pi}{6} \\ &\Rightarrow x=\tan \frac{\pi}{6} \\ &x=\frac{1}{\sqrt{3}} \end{aligned}$

Inverse Trigonometric Function Exercise 3.11 Question 3(v).

Answer:
$x=\sqrt{3}$
Hint:
Here, we use the formula of Intersection
$\tan^{-1}A-\tan^{-1}B=\tan^{-1}\left ( \frac{A-B}{1+AB} \right )$
Given:
$\cot ^{-1}x-\cot ^{-1}\left ( x+2 \right )=\frac{\pi }{12}$
Solution:
We know that,
$cot^{-1}x=\tan^{-1}\left ( \frac{1}{x} \right )$
Therefore, the given equation converted into
$\begin{aligned} &\tan ^{-1}\left(\frac{1}{x}\right)-\tan ^{-1}\left(\frac{1}{x+2}\right)=\frac{\pi}{12} \\ &\Rightarrow \tan ^{-1}\left(\frac{\frac{1}{x}-\frac{1}{x+2}}{1+\left(\frac{1}{x}\right)\left(\frac{1}{x+2}\right)}\right)=\frac{\pi}{12} \\ &\Rightarrow\left(\frac{\frac{x+2-x}{x(x+2)}}{\frac{x(x+2)+1}{x(x+2)}}\right)=\tan \frac{\pi}{12} \\ &\Rightarrow \frac{2}{x^{2}+2 x+1}=\tan \left(\frac{\pi}{3}-\frac{\pi}{4}\right) \end{aligned}$

$\Rightarrow \frac{2}{x^{2}+2x+1} = \tan \left ( \frac{\pi }{3} -\frac{\pi }{4}\right )$ $\left [ \because \frac{\pi }{3}-\frac{\pi }{4}= \frac{\pi }{12} \right ]$
$\begin{aligned} &\Rightarrow \frac{2}{x^{2}+2 x+1}=\frac{\tan \left(\frac{\pi}{3}\right)-\tan \left(\frac{\pi}{4}\right)}{1+\left(\tan \left(\frac{\pi}{3}\right)\right)\left(\tan \left(\frac{\pi}{4}\right)\right)} \\ &\Rightarrow \frac{2}{x^{2}+2 x+1}=\frac{\sqrt{3}-1}{1+\sqrt{3}} \end{aligned}$
$\left.\Rightarrow \frac{2}{x^{2}+2 x+1}=\frac{\sqrt{3}-1}{1+\sqrt{3}} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} \quad \text { multiply and divide by }(\sqrt{3}+1)\right]$
$\begin{aligned} &\Rightarrow \frac{2}{(x+1)^{2}}=\frac{3-1}{(1+\sqrt{3})^{2}} \\ &\Rightarrow \frac{2}{(x+1)^{2}}=\frac{2}{(1+\sqrt{3})^{2}} \\ &\Rightarrow(\sqrt{3}+1)^{2}=(x+1)^{2} \\ &\Rightarrow x+1=\sqrt{3}+1 \\ &\Rightarrow x=\sqrt{3} \end{aligned}$

Inverse Trigonometric Function Exercise 3.11 Question 3(vi)

Answer:
$x=\left (-20 \right ), \frac{1}{4}$
Hint:
Here, we use the formula
$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB} \right )$
Given:
$\tan^{-1}\left ( x+2 \right )+\tan^{-1}\left (x-2 \right )=\tan^{-1}\left ( \frac{8}{79} \right )$
Solution:
Here we have$A =x+2\: and\: B=x-2$
Using the value of A and B in the formula of $\tan^{-1}A+\tan^{-1}B$
$\begin{aligned} &\tan ^{-1}(x+2)+\tan ^{-1}(x-2)=\tan ^{-1}\left(\frac{8}{79}\right) \\ &\Rightarrow \tan ^{-1}\left[\frac{x+2+x-2}{1-(x+2)(x-2)}\right]=\tan ^{-1}\left(\frac{8}{79}\right) \\ &\Rightarrow\left[\frac{2 x}{1-x^{2}+4}\right]=\left(\frac{8}{79}\right) \end{aligned}$
$\begin{aligned} &\Leftrightarrow\left[\frac{2 x}{5-x^{2}}\right]=\left(\frac{8}{79}\right) \\ &\Rightarrow 2 x(79)=8\left(5-x^{2}\right) \\ &\Rightarrow 158 x=40-8 x^{2} \\ &\Rightarrow 8 x^{2}+158 x-40=0 \\ &\Rightarrow 79 x-20=0 \\ &\Rightarrow 4 x^{2}+80 x-x-20=0 \\ &\Rightarrow 4 x(x+20)-1(x+20)=0 \\ &\Rightarrow x+20=0 \text { or } 4 x-1=0 \\ &\Rightarrow x=-20 \text { or } 4 x=1 \end{aligned}$
$\Rightarrow x=\left (-20 \right ), 14$

Inverse Trigonometric Function Exercise 3.11 Question 3(vii)

Answer:
$x=1,\left ( -6 \right )$
Hint:
Here, we use the below formula
$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB} \right )$
Given:
$\tan^{-1}\left ( \frac{x}{2} \right )+\tan^{-1}\left ( \frac{x}{3} \right )=\frac{\pi }{4}$ for $0 < x < \sqrt{6}$
Solution:
$\! \! \! \! \! \! \! \! \! Here,A=\frac{x}{2}\\ B=\frac{x}{2}$
Using the values of A and B in the formula of $\tan^{-1}A+\tan^{-1}B$ , we get
$\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{x}{2}\right)+\tan ^{-1}\left(\frac{x}{3}\right)=\frac{\pi}{4} \\ &\Rightarrow \tan ^{-1}\left(\frac{\frac{x}{2}+\frac{x}{3}}{1-\left(\frac{x}{2}\right)\left(\frac{x}{3}\right)}\right)=\frac{\pi}{4} \\ &\Rightarrow \tan ^{-1}\left(\frac{\frac{3 x+2 x}{6}}{\frac{6-x^{2}}{6}}\right)=\frac{\pi}{4} \end{aligned}$
$\begin{aligned} &\Leftrightarrow \frac{5 x}{6-x^{2}}=\tan \frac{\pi}{4} \quad\left[\because \tan \frac{\pi}{4}=1\right] \\ &\Rightarrow 5 x=6-x^{2} \\ &\Rightarrow x^{2}+5 x-6=0 \\ &\Rightarrow x^{2}+6 x-x-6=0 \\ &\Rightarrow x(x+6)-1(x+6)=0 \\ &\Rightarrow(x-1)(x+6)=0 \end{aligned}$
$\! \! \! \! \! \! \! \! \! \Rightarrow Either x-1=0 \: or\: x+6=0 \\ \Rightarrow i\! f x-1=0\: then\: x=1 \\ i\! f x+6=0 \: then\: x=-6\\ \Rightarrow x=1,-6$

Inverse Trigonometric Function Exercise 3.11 Question 3(viii).:

Answer:
$x=\pm\sqrt{2}$
Hint:
Here, we will use the formula
$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB} \right )$
Given:
$\tan^{-1}\left ( \frac{x-2}{x-4} \right )+\tan^{-1}\left ( \frac{x+2}{x+4} \right )=\frac{\pi }{4}$
Solution:
$\begin{aligned} &\tan ^{-1}\left[\frac{\left(\frac{x-2}{x-4}\right)+\left(\frac{x+2}{x+4}\right)}{1-\left(\frac{x-2}{x-4}\right)\left(\frac{x+2}{x+4}\right)}\right]=\frac{\pi}{4} \\ &\Rightarrow \tan ^{-1}\left[\frac{\frac{x-2}{x-4}+\frac{x+2}{x+4}}{1-\left(\frac{x^{2}-2^{2}}{x^{2}-4^{2}}\right)}\right]=\frac{\pi}{4} \\ &\Rightarrow \frac{x^{2}+4 x-2 x-8+x^{2}-4 x+2 x-8}{x^{2}-16-x^{2}+4}=\tan \frac{\pi}{4} \\ &\Rightarrow \frac{2 x^{2}-16}{-12}=\tan \frac{\pi}{4} \\ &\Rightarrow \frac{2 x^{2}-16}{-12}=1 \quad\left[\because \tan \frac{\pi}{4}=1\right] \end{aligned}$
$\! \! \! \! \! \! \! \! \! \Rightarrow 2x^{2}-16=-12\\ \Rightarrow 2x^{2}=16-12\\ \Rightarrow 2x^{2}=4\\ \Rightarrow x^{2}=2\\ \Rightarrow x=\pm \sqrt2\\$

Inverse Trigonometric Function Exercise 3.11 Question 3(ix).

Answer:
$\Rightarrow x=\pm 3$
Hint:
Here, we will use the formula
$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB} \right )$
Given:
$\tan^{-1}\left ( 2+x \right )+\tan^{-1}\left ( 2-x\right )=\tan^{-1}\frac{2}{3}$
Solution:
$\begin{aligned} &\tan ^{-1}(2+x)+\tan ^{-1}(2-x)=\tan ^{-1} \frac{2}{3} \\ &\Rightarrow \tan ^{-1}\left(\frac{2+x+2-x}{1-(2+x)(2-x)}\right)=\tan ^{-1} \frac{2}{3} \\ &\Rightarrow \tan ^{-1}\left(\frac{4}{1-4+x^{2}}\right)=\tan ^{-1} \frac{2}{3} \\ &\Rightarrow \tan ^{-1}\left(\frac{4}{x^{2}-3}\right)=\tan ^{-1} \frac{2}{3} \end{aligned}$
$\! \! \! \! \! \! \! \! \! \Rightarrow 4x^{2}-3=23\\ \Rightarrow x^{2}-3=6\\ \Rightarrow x^{2}=9\\ \Rightarrow x=\pm 3\\$

Inverse Trigonometric Function Exercise 3.11 Question 3(x)

Answer:
$x=\pm \sqrt{\frac{7}{2}}$
Hint:
Here, we will use the formula
$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB} \right )$
Given:
$\tan^{-1}\left ( \frac{x-2}{x-1} \right )+\tan^{-1}\left ( \frac{x+2}{x+1} \right )=\frac{\pi }{4}$
Solution:
$\begin{aligned} &\Rightarrow \tan ^{-1}\left[\frac{\left(\frac{x-2}{x-1}\right)+\left(\frac{x+2}{x+1}\right)}{1-\left(\frac{x-2}{x-1}\right)\left(\frac{x+2}{x+1}\right)}\right]=\frac{\pi}{4}\\ &\Rightarrow \tan ^{-1}\left[\frac{\frac{x-2}{x-1}+\frac{x+2}{x+1}}{1-\left(\frac{x^{2}-2^{2}}{x^{2}-1^{2}}\right)}\right]=\frac{\pi}{4}\\ &\Rightarrow \tan ^{-1}\left[\frac{\frac{(x-2)(x+1)+(x+2)(x-1)}{x^{2}-1}}{\frac{x^{2}-1-x^{2}+4}{x^{2}-1}}\right]=\frac{\pi}{4}\\ &\Rightarrow\left[\frac{\frac{(x-2)(x+1)+(x+2)(x-1)}{x^{2}-1}}{\frac{x^{2}-1-x^{2}+4}{x^{2}-1}}\right]=\tan \frac{\pi}{4}\\ \end{aligned}$$\left [ \tan\frac{\pi }{4}= 1 \right ]$
$\! \! \! \! \! \! \! \! \! \Rightarrow\frac{x^{2}+x-2x-2+x^{2}-x+2x-2}{ 3}=1\\ \Rightarrow 2x^{2}-4=3\\ \Rightarrow 2x^{2}=7\\ \Rightarrow x^{2}=\frac{7}{2}\\\Rightarrow x= \pm \sqrt{\frac{7}{2}}$

Inverse Trigonometric Function Exercise 3.11 Question 3(xi).

Answer:
$x= \left ( \frac{1}{12} \right ),\left ( \frac{-1}{2} \right )$
Hint:
Here, we use the below formula
$\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB} \right )$
Given:
$\tan^{-1}4x+\tan^{-1}6x=\frac{\pi }{4}$
Solution:
Here A=4x
B=6x
Let’s put the values of A and B in formula of$\tan^{-1}A+\tan^{-1}B$

,we get
$\tan^{-1}\left ( \frac{4x+6x}{1-(4x)(6x)} \right )=\frac{\pi }{4}$
$\Rightarrow \tan^{-1}\left ( \frac{10x}{1-24x^{2}} \right )=\frac{\pi }{4}$
$\Rightarrow \frac{10x}{1-24x^{2}}=\tan \frac{\pi }{4}$
$\Rightarrow \frac{10x}{1-24x^{2}}=1$ $\left ( \tan \left ( \frac{\pi }{4} \right )= 1 \right )$
$\! \! \! \! \! \! \Rightarrow 10x=1-24x^{2}\\ \Rightarrow 24x^{2}+10x-1=0\\ \Rightarrow 24x^{2}+12x-2x-1=0\\ \Rightarrow 12x\left ( 2x+1 \right )-1\left ( 2x+1 \right )=0\\ \Rightarrow \left ( 12x-1 \right )\left ( 2x+1 \right )=0\\$$\! \! \! \! \! \! \! \! \! \Rightarrow Either\: 12x-1=0 \: or \: 2x+1=0\\ \Rightarrow 12x=1 \: or \: 2x= -1 \\ \Rightarrow x=\frac{1}{12} \: or \: x =-\frac{1}{2}\\ \Rightarrow x=\frac{1}{12},\left ( \frac{-1}{2} \right )\\$

Inverse Trigonometric Function Exercise 3.11 Question 4

Answer:
$\tan^{-1}2^{n}-\frac{\pi }{4}$
Hint:
We have to focus on calculation of infinite series.
Given:$\begin{aligned} &\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{2}{9}\right)+\ldots+\left.\tan ^{-1}\left(\frac{2^{n-1}}{1+2^{2 n-1}}\right)\right|_{-1} \\ \end{aligned}$

Solution:$\begin{aligned} &\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{2}{9}\right)+\ldots+\left.\tan ^{-1}\left(\frac{2^{n-1}}{1+2^{2 n-1}}\right)\right|_{-1} \\ &=\tan ^{-1}\left(\frac{2-1}{1+2 x 1}\right)+\tan ^{-1}\left(\frac{4-2}{1+4 x 2}\right)+\ldots+\tan ^{-1}\left(\frac{2^{n}-2^{n-1}}{1+2^{n} \cdot 2^{n-1}}\right) \\ &=\left(\tan ^{-1} 2-\tan ^{-1} 1\right)+\left(\tan ^{-1} 4-\tan ^{-1} 2\right)+\ldots .+\left(\tan ^{-1} 2^{n}-\tan ^{-1} 2^{n-1}\right) \\ &=\tan ^{-1} 2^{n}-\tan ^{-1} 1 \\ &=\tan ^{-1} 2^{n}-\frac{\pi}{4} \end{aligned}$

The RD Sharma class 12 chapter 3 exercise 3.11 book is based on the chapter Inverse Trigonometric Function. This chapter was also introduced in class 11, where students had learned the basic concepts of trigonometry. Exercise 3.11 in this chapter has a total of 16 questions, which are divided into levels 1 and 2. They are mostly addition and equation questions in trigonometry. The chapter on Inverse Trigonometric Functions is a critical chapter that needs special attention from students.

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Chapter-wise RD Sharma Class 12 Solutions