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KC Sinha Solution Class 12 Chapter 4 Inverse Trigonometric Function Exercise 4.1 Q1-Q5

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|>1
Sol :
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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