KC Sinha Solution Class 12 Chapter 4 Inverse Trigonometric Function Exercise 4.1 Q12-Q19

Exercise 4.1

Ex-4.1

Q1-Q5

Q6-Q11

Q12-Q19

TYPE-IV

Question 12
(i) $tan^{-1}x+tan^{-1}y+tan^{-1}z=\dfrac{\pi}{2}$
Prove that yz+zx+xy=1
Sol:

=$\tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{2}-\tan ^{2} z$

$=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\cot ^{-1} z$ $\left(\because \tan ^{-1} z+\cot \frac{-1}{z}=\frac{\pi}{2}\right)$

$=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\tan ^{-1} \frac{1}{z}$

=$\tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\cot ^{-1} z \quad\left(\because \tan ^{-1} z+\cot ^{-1}z=\frac{\pi}{2}\right)$

=$\tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\tan ^{-1} \frac{1}{z}$

$=\frac{x+y}{1-x y}=\frac{1}{z} \Rightarrow z(x+y)=1-x y$

$=z x+z y=1-x y$

$=y z+z x+x y=1$ proved

(ii) If $\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\pi,$ prove that

x+y+z = x y z
Sol :
Given $\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\pi$

let $\tan ^{-1} \alpha$ then $\tan \alpha=x$
$\tan ^{-1} y=\beta$ them $\tan \beta=y$

$\tan ^{-1} z=\gamma$ then $\tan \gamma=z$

Now, $\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\pi$

$=\tan ^{-1} x+\tan ^{-1} y=\pi-\tan ^{-1} z$

$=\gamma+\beta=\pi-y$

$=\tan (\alpha+\beta)=\tan (\pi-y)$

$\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}=-\tan y$

$=\frac{x+y}{1-x y}=-\frac{z}{1} \Rightarrow x+y=-z(1-x y)$

$=x+y=-z+x y z=x+y+z=x y z$ proved

Question 13

(i) If $\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}$, prove that

$x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=1$

Sol :

Given $\sin ^{-1} x+\sin ^{-1} y=\frac{x}{2}$

$=\sin ^{-1}(x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}})=\frac{x}{2}$

$=x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=\sin \frac{x}{2}$

$=x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=1$ proved

(ii) If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=x$, prove that
$x \sqrt{1-x^{2}}+y \sqrt{1-y^{2}}+z \sqrt{1-z^{2}}=2 x y z$

Sol :

Given $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=x$

let $\sin ^{-1} x=A$ then $\sin A=x$ $\therefore \cos A=\sqrt{1-x^{2}}$

$\sin ^{-1} y=B$ then $\sin B=y$ $\therefore \cos B=\sqrt{1-y^{2}}$

$\sin ^{-1} z=c \quad$ then $\quad \sin C=z \quad \therefore \cos C=\sqrt{1-z^{2}}$

Now, A+B+C=π

∴sin 2A+sin 2B+sin 2C=4 sin A.sin B.sin C

2sinA.cosA+2sinB.cosB+2sinC.cosC=4sinA.sinB.sinC

=2(sinA.cosA+sinB.cosB+sinC.cosC)=4sinA.sinB.sinC

=sinA.cosA+sinB.cosB+sinC.cosC=2sinA.sinB.sinC

=$x \cdot \sqrt{1-x^{2}}+y \cdot \sqrt{1-y^{2}}+z \cdot \sqrt{1-z^{2}}=2 \cdot x y z$ Proved

(iii) Establish the algebraic relation between x, y, z if $\tan ^{-1} x, \tan ^{-1} y, \tan ^{-1} z$ are in A.P and if further x, y, z are also in A.P. them prove that x=y=z

Sol:

Given $\tan ^{-1} x, \tan ^{-1} y, \tan ^{-1} z$ are in A.P

x,y,z are also in A.P

∴x-y=y-z ⇒x+z=2y...(i)

also , $\tan ^{-1} x+\tan ^{-1} z=2 \tan ^{-1} y$

$=\tan ^{-1}\left(\frac{x+z}{1-x z}\right)=\tan ^{-1} \frac{2 y}{1-y^{2}}$

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

$=\frac{2 y}{1-x z}=\frac{2 y}{1-y^{2}}=\frac{1}{1-x z}=\frac{1}{1-y^{2}}$

$=1-y^{2}=1-x z=-y^{2}=-x z$

$=y^{2}=x z$..(ii)

From equation (i) $\left(\dfrac{x+z}{2}\right)=y$

∴$y^2=\dfrac{(x+z)^2}{4}$...(iii)

Comparing equation (ii) and (iii) we get

xz=$\dfrac{(x+z)^2}{4}$⇒4xz=x²+z²+2xz

=x²+z²+2xz=0 ⇒(x-z)²=0

⇒(x-z)=0

⇒x=z..(iv)

Now $\frac{x+z}{1-x z}=\frac{2 y}{1-y^{2}} \Rightarrow \frac{2 y}{1-x^{2}}=\frac{2 y}{1-y^{2}}$

=1-x²=1-y²=x²=y²

x=y ..(v)

From equation (iv) and (v) we get

x=y=z Proved

Question 14

(i) $\cot ^{-1} x+\sin ^{-1} \frac{1}{\sqrt{5}}=\frac{\pi}{4}$

Sol :

$\Rightarrow \quad \tan ^{-1} \frac{1}{x}+\tan ^{-1} \frac{1}{2}=\frac{\pi}{4}$ $\left[\because \sin ^{-1} \frac{1}{\sqrt{5}}=\tan ^{-1} \frac{1}{2}\right]$

$\Rightarrow \frac{\left(\frac{1}{x}+\frac{1}{2}\right)}{\left(1-\frac{1}{x} \cdot \frac{1}{2}\right)}=\frac{x}{4}$

$\frac{\frac{2+x}{2 x}}{\frac{2x-1}{2 x}}=\tan \frac{\pi}{4}$

$\Rightarrow \frac{2+x}{2 x-1}=1 \Rightarrow 2+x=2 x-1$

=2x-x=3

⇒x=3

(ii) $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$
Sol :
Given $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$

$\Rightarrow \quad \sin ^{-1} \frac{1}{5}+\cos ^{-1} x=\sin ^{-1}(1)$

$\Rightarrow \quad \sin ^{-1} \frac{1}{5}+\cos ^{-1} x=\sin ^{-1} \sin \frac{\pi}{2}$

$\Rightarrow \quad \sin ^{-1} \frac{1}{5}+\cos ^{-1} x=\frac{\pi}{2}$

$\Rightarrow \quad \sin ^{-1} \frac{1}{5}=\frac{\pi}{2}-\cos ^{-1} x$

$\Rightarrow \sin \frac{-1}{5}=\sin ^{-1} x$

$\therefore x=\frac{1}{5} \quad$ Ans

Question 15

Solve 

(i) $\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$

Sol :

$\Rightarrow \tan ^{-1}\left(\frac{2 x+3 x}{1-2 x \cdot 3 x}\right)=\frac{\pi}{4}$

$\Rightarrow \frac{5 x}{1-6 x^{2}}=\tan \frac{\pi}{4} \Rightarrow \frac{5 x}{1-6 x^{2}}=1$

$\Rightarrow \quad 5 x=1-6 x^{2} \quad \Rightarrow \quad 6 x^{2}+5 x-1=0$

$\Rightarrow \quad 6 x^{2}+6 x-x-1=0$

$\Rightarrow \quad 6 x(x+1)-1(x+1)=0$

$\Rightarrow(x+1)(6 x-1)=0$

$x=-1, \quad x=\frac{1}{6} \quad$ Ans

(ii) $\tan ^{-1} x+\tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{\pi}{3}$

Sol :

$\Rightarrow \tan ^{-1} x+2 \tan ^{-1} x=\frac{\pi}{3}$

$\Rightarrow 3 \tan ^{-1} x=\frac{\pi}{3}$

$\Rightarrow \quad \tan ^{-1} x=\frac{\pi}{9}$


$\Rightarrow \quad x=\tan \frac{\pi}{9}$

(iii) $\tan ^{-1} \frac{1}{2}=\cot ^{-1} x+\tan ^{-1} \frac{1}{7}$

Sol :

$\tan ^{-1} \frac{1}{2}-\tan ^{-1} \frac{1}{7}=\tan^{-1} \frac{1}{x}$

⇒$\tan ^{-1} \frac{\left(\frac{1}{2}-\frac{1}{7}\right)}{\left(1+\frac{1}{2} \cdot \frac{1}{7}\right)}=\tan \frac{-1}{x}$

⇒$\tan ^{-1}\left(\frac{7-2}{14+1}\right)=\tan \frac{1}{x} \Rightarrow \frac{5}{15}=\frac{1}{x} \Rightarrow x=3$

Question 16

(i) $\tan ^{-1}(x-1)+\tan ^{-1} x+\tan ^{-1}(x+1)=\tan ^{-1} 3 x$
Sol :
$\Rightarrow \quad \tan ^{-1}(x-1)+\tan ^{-1}(x+1)=\tan ^{-1} 3 x-\tan ^{-1} x$

$\Rightarrow \tan ^{-1}\left(\frac{(x-1)+(x+1)}{1-(x-1)(x+1)}\right)=\tan ^{-1}\left(\frac{3 x-x}{1+3 x \cdot x}\right)$

$\Rightarrow \quad \frac{2 x}{1-\left(x^{2}-1\right)}=\frac{2 x}{1+3 x^{2}} \Rightarrow \frac{2 x}{2-x^{2}}=\frac{2 x}{1+3 x^{2}}$

$\begin{aligned} \Rightarrow 1+3 x^{2}=2-x^{2} & \Rightarrow 4 x^{2}=1 \\ & \Rightarrow x^{2}=\frac{1}{4} \end{aligned}$

$\Rightarrow x=\pm \dfrac{1}{2}$

(ii) $\tan ^{-1} \frac{x+1}{x-1}+\tan ^{-1} \frac{x-1}{x}=\pi+\tan ^{-1}(-7)$
Sol :
$\tan ^{-1} \frac{\left(\frac{x+1}{x-1}+\frac{x-1}{x}\right)}{1-\left(\frac{x+1}{x-1}\right)\left(\frac{x-1}{x}\right)}=\pi-\tan ^{-1} 7$

$\Rightarrow \tan ^{-1} \dfrac{\left(\frac{x(x+1)+(x-1)(x-1)}{x(x-1)}\right)}{\frac{x(x-1)-(x+1)(x-1)}{x(x-1)}}=\pi-\tan 7$

$\Rightarrow \tan ^{-1} \frac{\left[x(x+1)+(x-1)^{2}\right]}{x(x-1)-\left(x^{2}-1\right)}=\pi-\tan ^{-1} 7$

$\Rightarrow \frac{x(x+1)+(x-1)^{2}}{x(x-1)-\left(x^{2}-1\right)}=\tan \left(\pi-\tan ^{-1} 7\right)$

$\Rightarrow \frac{x^{2}+x+x^{2}+1-2 x}{x^{2}-x-x^{2}+1}=-\tan .\tan ^{-1} 7$

$\Rightarrow \frac{2 x^{2}-x+1}{1-x}=-7$ $(\because \tan (x-\theta)=-\tan\theta$

$\Rightarrow \quad 2 x^{2}-x+1=-7(1-x)$
$\Rightarrow 2 x^{2}-x+1=-7+7 x$
$\Rightarrow \quad 2 x^{2}-8 x+8=0$
$\Rightarrow \quad x^{2}-4 x+4=0$
$\Rightarrow x^{2}-2 x-2 x+4=0$
⇒x(x-2)-2(x-2)=0
⇒(x-2)(x-2)=0
⇒∴ x=2

(iii) $\tan ^{-1} \frac{x-1}{x-2}+\tan ^{-1} \frac{x+1}{x+2}=\frac{\pi}{4}$

Sol :

$\tan ^{-1} \frac{\left(\frac{x-1}{x-2}+\frac{x+1}{x+2}\right)}{1-\left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)}=\frac{\pi}{4}$

$\Rightarrow \frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x-1)(x+1)}=\tan \frac{\pi}{4}$

$\Rightarrow \frac{x^{2}+2 x-x-2+x^{2}-2 x+x-2}{x^{2}-4-\left(x^{2}-1\right)}=1$

$\Rightarrow \frac{2 x^{2}-4}{x^{2}-4-x^{2}+1} \Rightarrow \frac{2 x^{2}-4}{-3}=1$

$\Rightarrow \quad 2 x^{2}-4=-3 \quad \Rightarrow \quad 2 x^{2}=-3+4$

$\Rightarrow \quad 2 x^{2}=1 \quad \Rightarrow \quad x^{2}=\frac{1}{2}$

∴ $x=\pm \dfrac{1}{\sqrt{2}}$

Question 17

Solve

$\sin ^{-1} \frac{2 \alpha}{1+\alpha^{2}}+\sin ^{-1} \frac{2 \beta}{1+\beta^{2}}=2 \tan ^{-1} x|\alpha| \leq|\beta| \leq 1$

Sol :

$\Rightarrow \quad 2 \tan ^{-1} \alpha+2 \tan ^{-1} \beta=2 \tan ^{-1} x$

$\Rightarrow \quad 2\left(\tan ^{-1} \alpha+\tan ^{-1} \beta\right)=2 \tan ^{-1} x$

$\tan ^{-1} \alpha+\tan ^{-1} \beta=\tan ^{-1} x$

$\Rightarrow \quad \tan ^{-1}\left(\frac{\alpha+\beta}{1-\alpha \cdot \beta}\right)=\tan ^{-1} x$

$\Rightarrow \quad x=\frac{a+\beta}{1-\alpha \beta} \quad$ Ans

Question 18

Solve

(i) $\tan ^{-1} a x+\frac{1}{2} \sec ^{-1} b x=\frac{\pi}{4}$

Sol :

$\Rightarrow 2 \tan ^{-1}a x+\sec ^{-1} b x=\frac{\pi}{2}$

$\Rightarrow \tan ^{-1} \frac{2 a x}{1-a^{2} x^{2}}=\frac{x}{2}-\sec ^{-1} b x$

$\Rightarrow \frac{2 a x}{1-a^{2} x^{2}}=\tan \left(\frac{\pi}{2}-\sec ^{-1} b x\right)$

$\left[\because \tan \left(\frac{x}{2}-a\right)=\cot \theta \right]$

$\Rightarrow \frac{2 a x}{1-a^{2} x^{2}}=\tan \left(\frac{x}{2}-\cot ^{-1} \frac{1}{\sqrt{b^{2} x^{2}-1}}\right)$ 

$\Rightarrow \frac{2 a x}{1-a^{2} x^{2}}=\cot \cot ^{-1} \frac{1}{\sqrt{b^{2} x^{2}-1}}$

$\Rightarrow \frac{2 a x}{1-a^{2} x^{2}}=\frac{1}{\sqrt{b^{2} x^{2}-1}}$

$\Rightarrow 2ax\sqrt{b^{2} x^{2}-1}=\left(1-a^{2} x^{2}\right)$

$\Rightarrow 4 a^{2} x^{2}\left(b^{2} x^{2}-1\right)=\left(1-a^{2} x^{2}\right)$

$\Rightarrow 4 a^{2} b^{2} x^{4}-4 a^{2} x^{2}=1+a^{4} x^{4}-2 a^{2} x^{2}$

$\Rightarrow 4 a^{2} b^{2} x^{4}=1+a^{4} x^{4}-2 a^{2} x^{2}+4 a^{2} x^{2}$

$\Rightarrow 4 a^{2} b^{2} x^{4}=1+a^{4} x^{4}+2 a^{2} x^{2}$

$\Rightarrow \left(1+a^{2} x^{2}\right)^{2}=4 a^{2} b^{2} x^{4}$

$\Rightarrow 1+a^{2} x^{2}=\sqrt{4 a^{2}-b^{2} x^{4}}=2 a b x^{2}$

$\Rightarrow \quad 2 a b x^{2}-a^{2} x^{2}=1$

$\Rightarrow \quad x^{2} \times\left(2 a b-a^{2}\right)=1$

$\Rightarrow \quad x^{2}=\frac{1}{\left(2 a b-a^{2}\right)}$

$\Rightarrow \quad x=\pm \frac{1}{\sqrt{2 a b-a^{2}}}$ Ans

(iii) $\tan \left(\cos ^{-1} x\right)=\sin \left(\tan ^{-1} 2\right)$
Sol :
$\Rightarrow \tan \left(\tan ^{-1} \frac{\sqrt{1-x^{2}}}{x}\right)=\sin \left(\sin ^{-1} \frac{2}{\sqrt{5}}\right)$

$\Rightarrow \quad \frac{\sqrt{1-x^{2}}}{x}=\frac{2}{\sqrt{5}} \Rightarrow \frac{\left(1-x^{2}\right)}{x^{2}}=\frac{4}{5}$

$\Rightarrow \quad 5\left(1-x^{2}\right)=4 x^{2} \Rightarrow 5-5 x^{2}=4 x^{2}$

$\Rightarrow \quad 9 x^{2}=5$

$\Rightarrow \quad x^{2}=\frac{5}{9}$

$\Rightarrow \quad x=\sqrt{\frac{5}{9}}=\pm \frac{\sqrt{5}}{3}$


(iii) $\tan \left(\sec ^{-1} \frac{1}{x}\right)=\sin \cos ^{-1} \frac{1}{\sqrt{5}}$

Sol :

$\Rightarrow \tan \left(\cos ^{-1} x\right)=\sin \left(\sin ^{-1} \frac{2}{\sqrt{5}}\right)$

$\Rightarrow \tan \left(\tan \frac{\sqrt{1-x^{2}}}{x}\right)=\sin \left(\sin ^{-1} \frac{2}{\sqrt{5}}\right)$

$\Rightarrow \quad \frac{\sqrt{1-x^{2}}}{x}=\frac{2}{\sqrt{5}} \Rightarrow \frac{\left(1-x^{2}\right)}{x^{2}}=\frac{4}{5}$

$\Rightarrow \quad 5\left(1-x^{2}\right)=4 x^{2} \Rightarrow 5-5 x^{2}=4 x^{2}$

$\Rightarrow 4 x^{2}+5 x^{2}=5 \Rightarrow 9 x^{2}=5$

$\Rightarrow \quad x^{2}=\frac{5}{9}$

$\therefore x=\pm \frac{\sqrt{5}}{3}$ Ans

(iv) $\cos \left(\tan ^{-1} x\right)=\sin \left(\cot ^{-1} \frac{3}{4}\right)$
Sol :
$\Rightarrow \cos \left(\cos ^{1} \frac{1}{\sqrt{1+x^{2}}}\right)=\sin \left(\sin ^{-1} \frac{4}{5}\right)$

$\Rightarrow \frac{1}{\sqrt{1+x^{2}}}=\frac{4}{5} \Rightarrow \frac{1}{1+x^{2}}=\frac{16}{25}$

$\Rightarrow \quad 16\left(1+x^{2}\right)=25$

$\Rightarrow \quad 16+16 x^{2}=25$

$\Rightarrow \quad 16 x^{2}=25-16=9$


$16 x^{2}=9$

$\Rightarrow \quad x^{2}=\frac{9}{16}$

$\Rightarrow \quad x=\pm \frac{3}{4}$

Question 19

If $tan^{-1}\left(\frac{x-2}{x-4}\right)+\tan ^{-1}\left(\frac{x+2}{x+4}\right)=\frac{x}{4},$ Find the value of x

Sol :

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

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

$\Rightarrow \quad \frac{(x-2)(x+4)+(x+2)(x-4)}{(x-4)(x+4)-(x-2)(x+2)}=\tan \frac{\pi}{4}$

$\Rightarrow \frac{x^{2}+4 x-2 x-8+x^{2}-4 x+2 x-8}{\left(x^{2}-16\right)-\left(x^{2}-4\right)}=1$

$\Rightarrow \quad \frac{2 x^{2}-16}{x^{2}-16-x^{2}+4}=1$

⇒$\dfrac{2x^2-16}{-12}=1$

⇒2x²-12+16⇒2x²=4

⇒x²=2 $\Rightarrow x=\pm \sqrt{2}$

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