Exercise 4.1
| Ex-4.1 |
| Q1-Q5 |
| Q6-Q11 |
| Q12-Q19 |
TYPE 1
Question 1
Find the value of(i) $sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right)$
Sol :
Let $sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right)= \theta$..(i)
⇒$sin \theta =\dfrac{\sqrt{3}}{2}$
∵$\left[sin \dfrac{\pi}{3}=sin \dfrac{\sqrt{3}}{2}\right]$
⇒$sin \theta = sin \dfrac{\pi}{3}$
⇒$\theta =\dfrac{\pi}{3}$..(ii)
On comparing (i) and (ii) equations
⇒$sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right)=\dfrac{\pi}{3}$
(ii) $tan^{-1} \left(-\dfrac{1}{\sqrt{3}}\right)$
Sol :
⇒Let $tan^{-1} \left(-\dfrac{1}{\sqrt{3}}\right)= \theta$..(i)
⇒$tan \theta =-\dfrac{1}{\sqrt{3}}$
∵$tan \dfrac{\pi}{6}=sin \dfrac{1}{\sqrt{3}}$
⇒$tan \theta =-tan \dfrac{\pi}{6}$
⇒$tan \theta =tan \left(-\dfrac{\pi}{6}\right)$
⇒$\theta =-\dfrac{\pi}{6}$..(ii)
From (i) and (ii) , we get
⇒$tan^{-1}=-\dfrac{\pi}{6}$
(iii) $cot^{-1} (-\sqrt{3})$
Sol :
Note: cot-1 (-θ)=π-cot-1 θ
⇒cot-1 (-√3)=π-cot-1 (√3)
⇒$cot^{-1} (-\sqrt{3})=\pi -cot^{-1} \left(cot \dfrac{\pi}{6}\right)$
⇒$cot^{-1} (-\sqrt{3})=\pi-\dfrac{\pi}{6}$
⇒$cot^{-1} (-\sqrt{3})=\dfrac{6\pi-\pi}{6}=\dfrac{5\pi}{6}$
(vi) $cot^{-1}.cot \dfrac{5\pi}{4}$
Sol :
Let $cot^{-1} . cot\left(\dfrac{5\pi}{4}\right)=\theta$..(i)
⇒$cot \theta=cot \dfrac{5\pi}{4}$
⇒$cot \theta=cot \left(\pi +\dfrac{\pi}{4}\right)$
∵[cot(π+θ)=cotθ]
⇒$cot \theta =cot \dfrac{\pi}{4}$
⇒$\theta=\dfrac{\pi}{4}$..(ii)
From (i) and (ii) , we get
⇒$cot^{-1}. cot \dfrac{5\pi}{4}=\dfrac{\pi}{4}$
(v) $tan^{-1} \left(tan\dfrac{3\pi}{4}\right)$
Sol :
Let $tan^{-1} \left(tan\dfrac{3\pi}{4}\right)=\theta$..(i)
⇒$tan \theta = tan \dfrac{3\pi}{4}$
⇒$tan \theta = tan \left(\pi -\dfrac{\pi}{4}\right)$
[∵ tan(π-θ)=-tanθ]
⇒$tan \theta=-tan \dfrac{\pi}{4}$
⇒$tan \theta=tan \left(-\dfrac{\pi}{4}\right)$
⇒$\theta=-\dfrac{\pi}{4}$..(ii)
From (i) and (ii) , we get
⇒$tan^{-1}\left(tan \dfrac{3\pi}{4}\right)=-\dfrac{\pi}{4}$
(vi) $sin^{-1} \dfrac{1}{2}+cos^{-1} \dfrac{1}{2}$
Sol :
Note : $sin^{-1} x+ cos^{-1} x=\dfrac{\pi}{2}$
∴ $sin^{-1} \dfrac{1}{2} + cos^{-1} \dfrac{1}{2} =\dfrac{\pi}{2}$
(vii) $tan^{-1} \left(tan \dfrac{7\pi}{6}\right)$
Sol :
Let $tan^{-1}\left(tan \dfrac{7\pi}{6}\right)=\theta$..(i)
⇒$tan \theta=tan\dfrac{7\pi}{6}$
⇒$tan \theta=tan\left(\pi+\dfrac{\pi}{6}\right)$
[∵ tan(π+θ)=tanθ]
⇒$tan \theta = tan \dfrac{\pi}{6}$
⇒$\theta =\dfrac{\pi}{6}$..(ii)
From (i) and (ii) , we get
⇒$tan^{-1} \left(tan \dfrac{7\pi}{6}\right)=\dfrac{\pi}{6}$
(viii) $cos^{-1}. cos \left(\dfrac{13\pi}{6}\right)$
Sol :
Let $cos^{-1}. cos \left(\dfrac{13\pi}{6}\right)=\theta$..(i)
⇒$cos \theta =cos \dfrac{13\pi}{6}$
⇒$cos \theta =cos \left( \pi+\dfrac{7\pi}{6}\right)$
[∵ cos(π+θ)=-cosθ]
⇒$cos \theta=-cos\left(\dfrac{7\pi}{6}\right)$
⇒$cos \theta =-cos \left(\pi+\dfrac{\pi}{6}\right)$
⇒$cos \theta =cos \left(\dfrac{\pi}{6}\right)$
⇒$\theta=\dfrac{\pi}{6}$..(ii)
From (i) and (ii) , we get
⇒$cos^{-1}. cos \left(\dfrac{13\pi}{6}\right)=\dfrac{\pi}{6}$
(ix) $sin^{-1} \left(sin \dfrac{3\pi}{5}\right)$
Sol :
Let $sin^{-1} \left(sin \dfrac{3\pi}{5}\right)=\theta$..(i)
⇒$sin \theta =sin \dfrac{3\pi}{5}$
⇒$sin \theta =sin \left(\pi -\dfrac{2\pi}{5}\right)$
[∵ sin(π-θ)=sinθ]
⇒$sin \theta=sin \dfrac{2\pi}{5}$
⇒$\theta=\dfrac{2\pi}{5}$..(ii)
From (i) and (ii)
⇒$sin^{-1} \left(sin \dfrac{3\pi}{5}\right)=\dfrac{2\pi}{5}$
Question 2
(i) tan-1 (√3)Sol :
Let tan-1 (√3)=θ
⇒tanθ=√3
∵$\left[tan \dfrac{\pi}{3}=\sqrt{3}\right]$
⇒$tan \theta=tan \dfrac{\pi}{3}$
⇒$\theta =\dfrac{\pi}{3}$
(ii) $sin^{-1} \left(-\dfrac{1}{2}\right)$
Sol :
Let $sin^{-1} \left(-\dfrac{1}{2}\right)=\theta$
⇒$sin \theta = -\dfrac{1}{2}$
∵$\left[sin \dfrac{\pi}{6}=\dfrac{1}{2}\right]$
⇒$sin \theta =-sin \dfrac{\pi}{6}$
⇒$sin \theta =sin \left(-\dfrac{\pi}{6}\right)$
⇒$\theta=-\dfrac{\pi}{6}$
⇒$sin^{-1} \left(-\dfrac{1}{2}\right)=-\dfrac{\pi}{6}$
(iii) tan-1 (-1)
Sol :
Let tan-1 (-1)=θ
⇒tanθ=-1
∵$tan\dfrac{\pi}{4}=1$
⇒$\theta=-\dfrac{\pi}{4}$
⇒$tan^{-1} (-1)=-\dfrac{\pi}{4}$
(iv) cosec-1 (2)
Sol :
Let cosec-1 (2)=θ
⇒cosecθ=2
⇒$cosec \theta =cosec \dfrac{\pi}{6}$
⇒$\theta =\dfrac{\pi}{6}$
∴ $cosec^{-1} (2)=\dfrac{\pi}{6}$
(v) $cos^{-1} \left(-\dfrac{1}{2}\right)$
Sol :
Let $cos^{-1} \left(-\dfrac{1}{2}\right)=\theta$
⇒$cos \theta=-\dfrac{1}{2}$
⇒$cos \theta = cos \dfrac{2\pi}{3}$
∴$cos^{-1}=\dfrac{2\pi}{3}$
(vi) $cos^{-1} \left(\dfrac{\sqrt{3}}{2}\right)$
Sol :
Let $cos^{-1} \left(\dfrac{\sqrt{3}}{2}\right)=\theta$
⇒$cos \theta=\dfrac{\sqrt{3}}{2}$
⇒$cos \theta=\dfrac{\pi}{6}$
⇒$\theta=\dfrac{\pi}{6}$
∴ $cos^{-1} \left(\dfrac{\sqrt{3}}{2}\right)=\dfrac{\pi}{6}$
(vii) $cos^{-1} \left(cos \dfrac{2\pi}{3}\right)+sin^{-1} \left(sin \dfrac{2\pi}{3}\right)$
Sol :
Let $cos^{-1} \left(cos \dfrac{2\pi}{3}\right)=\alpha$ , $sin^{-1} \left(sin \dfrac{2\pi}{3}\right)=\beta$
⇒$cos \alpha =cos \dfrac{2\pi}{3}$ , $sin \beta = sin \dfrac{2\pi}{3}$
⇒$\alpha =\dfrac{2\pi}{3}$ ,
$sin \beta =sin \left(\pi -\dfrac{\pi}{3}\right)$
[∵ sin(π-θ)=sinθ]
$sin \beta = sin\dfrac{\pi}{3}$
$\beta = \dfrac{\pi}{3}$
⇒$cos^{-1} \left(cos \dfrac{2\pi}{3}\right)+sin^{-1} \left(sin \dfrac{2\pi}{3}\right)$
⇒α+β
⇒$\dfrac{2\pi}{3}+\dfrac{\pi}{3}=\dfrac{3\pi}{3}=\pi$
(viii) $cos^{-1} \left(\dfrac{1}{2}\right)-2sin^{-1} \left(-\dfrac{1}{2}\right)$
Sol :
[∵ sin-1 (-x)=-sin-1 x]
⇒$cos^{-1} \left(cos \dfrac{\pi}{3}\right)+2sin^{-1} \left(\dfrac{1}{2}\right)$
⇒$cos^{-1} \left(cos \dfrac{\pi}{3}\right)+2sin^{-1} \left(sin\dfrac{\pi}{6}\right)$
⇒$\dfrac{\pi}{3}+\dfrac{2\pi}{6}$
⇒$\dfrac{\pi}{3}+\dfrac{\pi}{3}=\dfrac{2\pi}{3}$
Question 3
(i) $cos \left[tan^{-1}\left(\dfrac{3}{4}\right)\right]$Sol :
Let $tan^{-1}\dfrac{3}{4}=\theta$
⇒$tan \theta =\dfrac{3}{4}=\dfrac{p}{b}$
[h=√p2+b2
=√32+42
=√25=5]
⇒$cos \theta =\dfrac{b}{h}=\dfrac{4}{5}$
⇒$ \theta =cos^{-1} \dfrac{4}{5}$
⇒$tan^{-1}\dfrac{3}{4}=cos^{-1}\dfrac{4}{5}$
On putting $cos^{-1} \dfrac{4}{5}$ in the place of $tan^{-1}\dfrac{3}{4}$
We get $cos \left[cos^{-1}\dfrac{4}{5}\right]=\dfrac{4}{5}$
(ii) $cos \left[cos^{-1} \left(\dfrac{\sqrt{3}}{2}\right)+\dfrac{\pi}{6}\right]$
Sol :
⇒$cos \left[\dfrac{\pi}{6}+\dfrac{\pi}{6}\right]$
⇒$cos \left[\dfrac{2\pi}{6}\right]=cos \dfrac{\pi}{3}=\dfrac{1}{2}$
(iii) $2arc~sin\left(\dfrac{1}{2}\right)+3arc~tan (-1)+2arc~cos\left(-\dfrac{1}{2}\right)$
Sol :
⇒$2sin^{-1} \left(\dfrac{1}{2}\right)+3tan^{-1} (-1)+2cos^{-1} \left(-\dfrac{1}{2}\right)$
⇒$2sin^{-1}\left(sin \dfrac{\pi}{6}\right)-3tan^{-1} (1)+2cos^{-1} \left(cos \dfrac{2\pi}{3}\right)$
⇒$2sin^{-1}\left(sin \dfrac{\pi}{6}\right)-3tan^{-1} \left(tan \dfrac{\pi}{4}\right)+2cos^{-1} \left(cos \dfrac{2\pi}{3}\right)$
⇒$2\times \dfrac{\pi}{6}-\dfrac{3\pi}{4}+2\times \dfrac{2\pi}{3}$
⇒$\dfrac{\pi}{3}-\dfrac{3\pi}{4}+\dfrac{4\pi}{3}=\dfrac{4\pi-9\pi+16\pi}{12}$
⇒$\dfrac{20\pi-9\pi}{12}=\dfrac{11\pi}{12}$
TYPE 2
Question 4
(i) $tan^{-1} \left(\dfrac{1}{\sqrt{x^2-1}}\right)$ |x|>1Sol :
Putting x=secθ , then θ=sec-1 x
Now , $tan^{-1} \left(\dfrac{1}{\sqrt{sec^2 \theta-1}}\right)$
⇒$tan^{-1} \left(\dfrac{1}{\sqrt{tan^{2} \theta}}\right)$
⇒$tan^{-1} \left(\dfrac{1}{tan \theta}\right)$
⇒tan-1 cotθ
⇒$tan^{-1} . tan \left(\dfrac{\pi}{2}-\theta\right)$
⇒$\dfrac{\pi}{2}-\theta=\dfrac{\pi}{2}-sec^{-1} x$
(ii) $tan^{-1} \dfrac{\sqrt{1+x^2}-1}{2}$ , x≠0
Sol :
Putting x=tanθ and θ=tan-1 x
Now $tan^{-1} \dfrac{\sqrt{1+tan^2 \theta}-1}{tan \theta}$
⇒$tan^{-1} \dfrac{\sqrt{sec^2 \theta}-1}{tan \theta}$
⇒$tan^{-1} \dfrac{sec \theta -1}{tan \theta}$
⇒$tan^{-1} \dfrac{\dfrac{1}{cos \theta}-1}{\dfrac{sin \theta}{cos \theta}}$
⇒$tan^{-1} \dfrac{1-cos \theta}{cos \theta} \times \dfrac{cos \theta}{sin \theta}$
⇒$tan^{-1} \dfrac{1- cos \theta}{sin \theta}$
∵$\left[1-cos \theta=2sin^2 \dfrac{\theta}{2}\right]$
∵$sin \theta =2sin \dfrac{\theta}{2}.cos \dfrac{\theta}{2}$
⇒$tan^{-1} \dfrac{2sin^2 \dfrac{\theta}{2}}{2sin \dfrac{\theta}{2}.cos \dfrac{\theta}{2} }$
⇒$tan^{-1} \dfrac{sin \dfrac{\theta}{2}}{cos \dfrac{\theta}{2}}$
⇒$tan^{-1} tan \dfrac{\theta}{2}=\dfrac{\theta}{2}$
⇒$\dfrac{tan^{-1} x}{2}$
(iii) $tan^{-1} \dfrac{cos x}{1+sinx}$
Sol :
Note: $cos x=\dfrac{1-tan^2 \dfrac{x}{2}}{1+tan^2 \dfrac{x}{2}}$
$sin x=\dfrac{2tan \dfrac{x}{2}}{1+tan^2 \dfrac{x}{2}}$
⇒$tan^{-1}\dfrac{\left(\dfrac{1-tan^2 \frac{x}{2}}{1+tan^2 \frac{x}{2}}\right)}{\left(1+\dfrac{2tan \frac{x}{2}}{1+tan^2 \frac{x}{2}}\right)}$
⇒$tan^{-1} \dfrac{\left(\dfrac{1-tan^2 \frac{x}{2}}{1+tan^2 \frac{x}{2}}\right)}{\left(\dfrac{1+tan^2 \frac{x}{2}+2tan \frac{x}{2}}{1+tan^2 \frac{x}{2}}\right)}$
⇒$tan^{-1} \left[\dfrac{1^2-tan^2 \frac{x}{2}}{\left(1+tan \frac{x}{2}\right)^2}\right]$
⇒$tan^{-1} \dfrac{\left(1-tan \frac{x}{2}\right)\left(1+tan \frac{x}{2}\right)}{\left(1+tan \frac{x}{2}\right)\times\left(1+tan \frac{x}{2}\right)}$
⇒$tan^{-1} \left(\dfrac{1-tan \frac{x}{2}}{1+tan \frac{x}{2}}\right)$
⇒$tan^{-1} \left(\dfrac{tan\frac{\pi}{4}-tan \frac{x}{2}}{1+tan\frac{\pi}{4}.tan \frac{x}{2}}\right)$
$\left[tan (A-B)=\dfrac{tanA-tanB}{1+tanA.tanB}\right]$
⇒$tan^{-1} . tan\left(\dfrac{x}{4}-\dfrac{x}{2}\right)$
⇒$\dfrac{x}{4}-\dfrac{x}{2}$
(iv) $cot^{-1} \left[\dfrac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\right]$
Sol :
⇒$cot^{-1}\left[\dfrac{1+\dfrac{\sqrt{1-sinx}}{\sqrt{1+sinx}}}{1-\dfrac{\sqrt{1-sinx}}{\sqrt{1+sinx}}}\right]$
Note : $1-cos \theta =2sin^2 \dfrac{\theta}{2}$ $1+cos \theta =2cos^2 \dfrac{\theta}{2}$
Now $\sqrt{\dfrac{1-sinx}{1+sinx}}=\sqrt{\dfrac{1-cos\left(\frac{x}{2}-x\right)}{1+cos\left(\frac{x}{2}-x\right)}}$ $\sqrt{\dfrac{2 sin^2 \left(\frac{\pi}{4}-\frac{x}{2}\right)}{2cos^2 \left(\frac{\pi}{4}-\frac{x}{2}\right)}}$
⇒$\dfrac{sin\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)}{cos\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)}=tan \left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$
∴$\sqrt{\dfrac{1-sin x}{1+sin x }}=tan \left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$
⇒$cot^{-1}\left(\dfrac{1+tan\left(\frac{\pi}{4}-\frac{x}{2}\right)}{1-tan \left(\frac{\pi}{4}-\frac{x}{2}\right)}\right)$
∵$\dfrac{tan A+tanB}{1-tanA.tanB}=tan(A+B)$
⇒$cot^{-1}\left(\dfrac{tan\frac{\pi}{4}+tan\left(\frac{\pi}{4}-\frac{x}{2}\right)}{1-tan \frac{\pi}{4}.tan\left(\frac{\pi}{4}-\frac{x}{2}\right)}\right)$
⇒$cot^{-1} . tan\left(\dfrac{\pi}{4}+\dfrac{\pi}{4}-\dfrac{x}{2}\right)$
⇒$cot^{-1} cot \left(\dfrac{\pi}{2}-\left(\dfrac{\pi}{4}+\dfrac{\pi}{4}-\dfrac{x}{2}\right)\right)$
⇒$\dfrac{\pi}{2}-\dfrac{\pi}{4}-\dfrac{\pi}{4}+\dfrac{x}{2}=\dfrac{x}{2}$
(v) $cos^{-1} \sqrt{\dfrac{\sqrt{1+x^2}+1}{2\sqrt{1+x^2}}}$
Sol :
Putting x=tanθ , then θ=tan-1 x
Now $cos^{-1} \sqrt{\dfrac{\sqrt{1+tan^2 \theta}+1}{2\sqrt{1+tan^2 \theta}}}$
[∵ 1+tan2θ=sec2θ]
⇒$cos^{-1} \sqrt{\dfrac{sec\theta+1}{2 sec \theta}}$
⇒$cos^{-1} \sqrt{\dfrac{1+cos \theta}{\dfrac{2}{cos \theta}\times cos \theta}}$
⇒$cos^{-1} \sqrt{\dfrac{1+cos \theta}{2}}$
⇒$cos^{-1} \sqrt{\dfrac{2cos^2 \frac{\theta}{2}}{2}}$
⇒$cos^{-1} . cos\dfrac{\theta}{2}$
⇒$\dfrac{\theta}{2}=\dfrac{tan^{-1}x}{2}$
(vi) $tan^{-1} \left(\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$
Sol :
Putting x=cosθ , then $\theta=\dfrac{1}{2}cos^{-1}x$
Now , $tan^{-1}\left(\dfrac{\sqrt{1+cos2\theta}-\sqrt{1-cos2\theta}}{\sqrt{1+cos2\theta}+\sqrt{1-cos2\theta}}\right)$
⇒$tan^{-1}\left(\dfrac{\sqrt{2cos^2\theta}-\sqrt{2sin^2\theta}}{\sqrt{2cos^2\theta}+\sqrt{2sin^2\theta}}\right)$
⇒$tan^{-1} \dfrac{\sqrt{2}cos \theta-\sqrt{2}sin \theta}{\sqrt{2}cos \theta+\sqrt{2}sin \theta}$
⇒$tan^{-1} \dfrac{\sqrt{2}(cos \theta-sin \theta)}{\sqrt{2}(cos \theta+sin \theta)}$
[dividing by cosθ]
⇒$tan^{-1}\left(\dfrac{1-tan\theta}{1+tan \theta}\right)$
⇒$tan^{-1}\left(\dfrac{tan \dfrac{\pi}{4}-tan\theta}{1+tan \dfrac{\pi}{4} .tan \theta}\right)$
⇒$tan^{-1} tan \left(\dfrac{\pi}{4}-\theta\right)$
⇒$\dfrac{\pi}{4}-\theta$
⇒$\dfrac{\pi}{4}-\dfrac{1}{2} cos^{-1}x$
TYPE 3
Question 5
Prove that:(i) $tan^{-1} \dfrac{2}{11}+tan^{-1} \dfrac{7}{24}$
Sol :
Note: $tan^{-1} A+tan^{-1} B=tan^{-1}\left(\dfrac{A+B}{1-AB}\right)$
Now , L.H.S = $tan^{-1} \dfrac{2}{11}+tan^{-1} \dfrac{7}{24}=tan^{-1} \left(\dfrac{\dfrac{2}{11}+\dfrac{7}{24}}{1-\dfrac{2}{11}\times \dfrac{7}{24}}\right)$
⇒$tan^{-1} \left(\dfrac{\dfrac{48+77}{264}}{1-\dfrac{14}{264}}\right)$
⇒$tan^{-1} \left(\dfrac{125}{250}\right)=tan^{-1} \left(\dfrac{1}{2}\right)$ =R.H.S
Hence Proved
(ii) $tan^{-1} \dfrac{1}{2}+tan^{-1} \dfrac{1}{3}=tan^{-1} \dfrac{3}{5}+tan^{-1} \dfrac{1}{4}=\dfrac{\pi}{4}$
Sol :
Note : $tan^{-1} A + tan^{-1} B = tan^{-1} \left(\dfrac{A+B}{1-AB}\right)$
Now L.H.S=$tan^{-1}\dfrac{1}{2}+tan^{-1}\dfrac{1}{3}=tan^{-1}\left(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}}{1-\dfrac{1}{2}\times \dfrac{1}{3}}\right)$
⇒$tan^{-1} \left(\dfrac{\dfrac{3+2}{6}}{\dfrac{6-1}{6}}\right)$
⇒$tan^{-1} \left(\dfrac{5}{5}\right)=tan^{-1} (1)$
⇒$tan^{-1} \left(tan \dfrac{\pi}{4}\right)=\dfrac{\pi}{4}$ R.H.S..(i)
Again , L.H.S=$tan^{-1} \dfrac{3}{5}+tan^{-1} \dfrac{1}{4}=tan^{-1} \left(\dfrac{\dfrac{3}{5}+\dfrac{1}{4}}{1-\dfrac{3}{5}\times \dfrac{1}{4}}\right)$
⇒$tan^{-1} \left(\dfrac{\dfrac{12+5}{20}}{1-\dfrac{3}{20}}\right)$
⇒$tan^{-1} \left(\dfrac{\dfrac{17}{20}}{\dfrac{20-3}{20}}\right)$
⇒$tan^{-1} \left(\dfrac{17}{17}\right)=tan^{-1} (1)$
⇒$tan^{-1} \left(tan \dfrac{\pi}{4}\right)=\dfrac{\pi}{4}$..(ii)
From (i) and (ii) , we get
$tan^{-1} \dfrac{1}{2}+tan^{-1} \dfrac{1}{3}=tan^{-1} \dfrac{3}{5}+tan^{-1} \dfrac{1}{4}=\dfrac{\pi}{4}$
Hence Proved
(iii) $tan^{-1} \dfrac{2a-b}{b\sqrt{3}}+tan^{-1} \dfrac{2b-a}{a\sqrt{3}}=\dfrac{\pi}{3}$
Sol :
L.H.S =$tan^{-1} \dfrac{2a-b}{b\sqrt{3}}+tan^{-1} \dfrac{2b-a}{a\sqrt{3}}=\dfrac{\pi}{3}$
⇒$tan^{-1} \left(\dfrac{\dfrac{2a-b}{b\sqrt{3}}+\dfrac{2b-a}{a\sqrt{3}}}{1-\dfrac{2a-b}{b\sqrt{3}}\times \dfrac{2b-a}{a\sqrt{3}}}\right)$
⇒$tan^{-1} \left(\dfrac{\dfrac{a\sqrt{3}(2a-b)+b\sqrt{3}(2b-a)}{b\sqrt{3}.a\sqrt{3}}}{1-\dfrac{(2a-b)(2b-a)}{b\sqrt{3}.a\sqrt{3}}}\right)$
⇒$tan^{-1} \left(\dfrac{\dfrac{2\sqrt{3}a^2-ab\sqrt{3}+2\sqrt{3}b^2-ab\sqrt{3}}{b\sqrt{3}.a\sqrt{3}}}{\dfrac{b\sqrt{3}.a\sqrt{3}-(4ab-2a^2-2b^2+ab)}{b\sqrt{3}.a\sqrt{3}}}\right)$
⇒$tan^{-1} \left(\dfrac{2\sqrt{3}a^2+2\sqrt{3}b^2-2ab\sqrt{3}}{3ab-4ab+2a^2+2b^2-ab}\right)$
⇒$tan^{-1} \left(\dfrac{2\sqrt{3}(a^2+b^2-ab)}{2(a^2+b^2-ab)}\right)$
⇒$tan^{-1} (\sqrt{3})=tan^{-1} (tan \dfrac{\pi}{3})=\dfrac{\pi}{3}$ =R.H.S
Hence Proved
(iv) $tan^{-1} \dfrac{2}{5}+tan^{-1} \dfrac{1}{3}+tan^{-1} \dfrac{1}{12}=\dfrac{\pi}{4}$
Sol :
L.H.S=$tan^{-1} \dfrac{2}{5}+tan^{-1} \dfrac{1}{3}+tan^{-1} \dfrac{1}{12}$
⇒$tan^{-1} \left(\dfrac{\frac{2}{5}+\frac{1}{3}}{1-\frac{2}{5} \times \frac{1}{3}}\right)+tan^{-1} \dfrac{1}{12}$
⇒$tan^{-1} \left(\dfrac{\dfrac{6+5}{15}}{\dfrac{15-2}{15}}\right)+tan^{-1} \dfrac{1}{12}$
⇒$tan^{-1} \left(\dfrac{11}{13}\right)+tan^{-1} \dfrac{1}{12}$
⇒$tan^{-1} \left(\dfrac{\frac{11}{13}+\frac{1}{12}}{1-\frac{11}{13}\times \frac{1}{12}}\right)$
⇒$tan^{-1} \left(\dfrac{\frac{132+13}{156}}{\frac{156-11}{156}}\right)$
⇒$tan^{-1} \left(\dfrac{145}{145}\right)=tan^{-1}$
⇒$tan^{-1} \left(tan\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}$=R.H.S proved
(v) $2tan^{-1} \dfrac{1}{5}+tan^{-1} \dfrac{1}{4}=tan^{1} \dfrac{32}{45}$
Sol :
∵$\left(2tan^{-1} x=tan^{-1}\dfrac{2x}{1-x^2}\right)$
⇒$2tan^{-1} \dfrac{1}{5}=tan^{-1} \left(\dfrac{2\dfrac{1}{5}}{1-\dfrac{1}{5^2}}\right)$
⇒$tan^{-1} \dfrac{\dfrac{2}{5}}{\dfrac{25-1}{25}}=tan^{-1} \dfrac{10}{24}$
⇒$tan^{-1} \dfrac{5}{12}$
∴$2tan^{-1} \dfrac{1}{5}=tan^{-1} \dfrac{5}{12}$
Now, L.H.S $2tan^{-1} \dfrac{1}{5}+tan^{-1} \dfrac{1}{4}$
⇒$tan^{-1} \dfrac{5}{12}+tan^{-1} \dfrac{1}{4}$
⇒$tan^{-1} \left(\dfrac{\frac{5}{12}+\frac{1}{4}}{1-\frac{5}{12}\times \frac{1}{4}}\right)$
⇒$tan^{-1} \left(\dfrac{\frac{20+12}{48}}{\frac{48-5}{48}}\right)$
⇒$tan^{-1} \left(\dfrac{32}{43}\right)$ R.H.S proved
(vi) $2tan^{-1} \dfrac{1}{2}+tan^{-1} \dfrac{1}{7}=tan^{-1} \dfrac{31}{17}$
Sol :
Note :$\left(2tan^{-1} x=tan^{-1}\dfrac{2x}{1-x^2}\right)$
⇒$2 tan^{-1} \dfrac{1}{2}=tan^{-1} \dfrac{2\times \frac{1}{2}}{1-\frac{1}{2^2}}$
$=tan^{-1} \dfrac{1}{1-\frac{1}{4}}=tan^{-1} \dfrac{4}{4-1}$
$=tan^{-1} \dfrac{4}{3}$
Now , L.H.S =$2tan^{-1} \dfrac{1}{2}+tan^{-1} \dfrac{1}{7}$
⇒$tan^{-1} \dfrac{4}{3}+tan^{-1} \dfrac{1}{7}$
⇒$tan^{-1} \left(\dfrac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}\times \frac{1}{7}}\right)$
⇒$tan^{-1} \left(\dfrac{\frac{28+3}{21}}{\frac{21-4}{21}}\right)$
⇒$tan^{-1} \dfrac{31}{17}$ R.H.S proved
(vii) tan-1 1+tan-1 2+tan-1 3=π$=2(tan^{-1} 1+tan^{-1}\dfrac{1}{2}+tan^{-1}\dfrac{1}{3})$
Sol :
L.H.S =tan-1 1+tan-1 2+tan-1 3
=$tan^{-1} \left(\dfrac{1+2}{1-1\times2}\right)+tan^{-1}3$
=$tan^{-1} \left(\dfrac{3}{1-2}\right)+tan^{-1}3$
=$tan^{-1} \left(\dfrac{3}{-1}\right)+tan^{-1}3$
=tan-1 3 - tan-1 3
=$tan^{-1} \left(\dfrac{3-3}{1+3\times 3}\right)$
=$tan^{-1} \left(\dfrac{0}{10}\right)=tan^{-1}0$
=tan-1(tanπ)=π=R.H.S..(i)
Again , R.H.S=$2\left(tan^{-1} 1+tan^{-1} \dfrac{1}{2}+tan^{-1} \dfrac{1}{3}\right)$
=$2\left\{tan^{-1}\left(\dfrac{1+\frac{1}{2}}{1-1\times \frac{1}{2}}\right)+tan^{-1} \dfrac{1}{3}\right\}$
=$2\left\{tan^{-1}\left(\dfrac{\frac{3}{2}}{\times \frac{2-1}{2}}\right)+tan^{-1} \dfrac{1}{3}\right\}$
=$2 \left\{tan^{-1} \dfrac{3}{1}+tan^{-1} \dfrac{1}{3}\right\}$
=$2 \left\{tan^{-1} \left(\dfrac{3+\frac{1}{3}}{1-3\times \frac{1}{3}}\right) \right\}$
=$2 \left\{tan^{-1} \left(\dfrac{\frac{10}{3}}{\frac{1-1}{3}}\right) \right\}$
=$2 \left\{tan^{-1} \left(\dfrac{10}{0}\right) \right\}$
=$2 \left\{ tan^{-1} (undefined)\right\}=2\left\{tan^{-1} tan \dfrac{\pi}{2}\right\}$
=$2 \dfrac{\pi}{2}=\pi$ L.H.S ..(ii)
From (i) and (ii)
⇒tan-1 1+tan-1 2+tan-1 3=π$=2(tan^{-1} 1+tan^{-1}\dfrac{1}{2}+tan^{-1}\dfrac{1}{3})$
(viii) $tan^{-1} \left[2cos \left(2sin^{-1} \dfrac{1}{2}\right)\right]$
Sol :
Given , $tan^{-1} \left[2cos \left(2sin^{-1} \dfrac{1}{2}\right)\right]$
=$tan^{-1} \left[2cos (2sin^{-1} sin \dfrac{\pi}{6})\right]$
=$tan^{-1} \left[2cos \left(2 \dfrac{\pi}{6}\right)\right]$
=$tan^{-1} \left[2cos \dfrac{\pi}{3}\right] =\tan^{-1} \left[2\times \dfrac{1}{2}\right]$
=$tan^{-1} (1)=tan^{-1} \left(tan \dfrac{\pi}{4}\right)=\dfrac{\pi}{4}$ R.H.S
proved
(ix) $tan^{-1} \left(sin^{-} \dfrac{3}{5}+cot^{-1} \dfrac{3}{2}\right)=\dfrac{17}{6}$
Sol :
Let $sin^{-1} \dfrac{3}{5}=\alpha$ and $cot^{-1} \dfrac{3}{2}=\beta$
⇒$sin \alpha =\dfrac{3}{5} , cot \beta =\dfrac{3}{2}$
⇒$tan \alpha =\dfrac{3}{4} , tan \beta =\dfrac{2}{3}$
⇒$\alpha=tan^{-1} \dfrac{3}{4} , \beta=tan^{-1} \dfrac{2}{3}$
⇒$sin^{-1} \dfrac{3}{5}=tan^{-1} \dfrac{3}{4} , cot^{-1} \dfrac{3}{2}=tan^{-1} \dfrac{2}{3}$
Now , L.H.S=$tan \left(sin^{-1} \dfrac{3}{5}+cot^{-1} \dfrac{3}{2}\right)$
$=tan\left(tan^{-1} \dfrac{3}{4}+tan^{-1} \dfrac{2}{3}\right)$
$=tan \left(tan^{-1} \left(\dfrac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\times \frac{2}{3}}\right)\right)$
$=tan \left(tan^{-1} \left(\dfrac{\frac{9+8}{12}}{\frac{12-6}{12}}\right)\right)$
=$tan \left(tan^{-1} \dfrac{17}{6}\right)=\dfrac{17}{6}$ R.H.S
proved
(x) $tan^{-1} \left(\dfrac{1}{2} sin^{-1} \dfrac{3}{4}\right)=\dfrac{4-\sqrt{7}}{3}$
Sol :
Let $\left(\dfrac{1}{2} sin^{-1} \dfrac{3}{4}\right)=\theta$
⇒$sin^{-1} \dfrac{3}{4}=2\theta$
⇒$sin^{-1} 2\theta=\dfrac{3}{4}$
⇒$\dfrac{2tan \theta}{1+tan^{2}\theta}=\dfrac{3}{4}$
⇒8 tanθ=3(1+tan2 θ)
⇒8 tanθ=3+3tan2 θ)
⇒3tan2 θ-8tanθ+3=0
Let tanθ=x , then 3x2-8x+3=0
Discriminant (D)=b2-4ac
=(-8)2-4×3×3
=64-36=28>0
$x=\dfrac{-b \pm \sqrt{D}}{2a}=\dfrac{-(-8) \pm \sqrt{28}}{2\times3}$
$x=\dfrac{8 \pm \sqrt{4\times 7}}{2\times 3}=\dfrac{8 \pm 2\sqrt{7}}{2\times 3}$
$x=\dfrac{2(4 \pm \sqrt{7})}{2\times 3}=\dfrac{4 \pm \sqrt{7}}{3}$
$x=\dfrac{4+\sqrt{7}}{3}$ or $x=\dfrac{4-\sqrt{7}}{3}$
⇒$tan\theta=\dfrac{4-\sqrt{7}}{3}$
⇒$\theta=tan^{-1}\left(\dfrac{4-\sqrt{7}}{3}\right)$
⇒$\dfrac{1}{2} sin^{-1} \dfrac{3}{4}=\theta=tan^{-1} \left(\dfrac{4-\sqrt{7}}{3}\right)$
⇒$tan\left(tan^{-1} \left(\dfrac{4-\sqrt{7}}{3}\right)\right)=\dfrac{4-\sqrt{7}}{3}$ R.H.S
Proved
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