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

Exercise 4.1


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


Question 6

Prove that

(i) tan^{-1} x+cot^{-1} y=tan^{-1} \dfrac{xy+1}{y-x}

Sol :

L.H.S=tan-1 x + cot-1 y

\left[cot^{-1} x =tan^{-1} \dfrac{1}{x}\right]

=tan^{-1} x+tan^{-1}\dfrac{1}{y}

=tan^{-1} \left(\dfrac{x+\dfrac{1}{y}}{1-x\times \dfrac{1}{y}}\right)

=tan^{-1} \left( \dfrac{\dfrac{xy+1}{y}}{\dfrac{y-x}{y}} \right)

=tan^{-1} \dfrac{xy+1}{y-x} R.H.S


(ii) tan-1 x + cot-1(1+x) = tan-1(1+x+x2)

Sol :

L.H.S= tan-1 x+cot-1 (1+x)

=tan^{-1} x+tan^{-1} \dfrac{1}{1+x}

=tan^{-1} \left(\dfrac{x+\dfrac{1}{x}}{1-x\times \dfrac{1}{1+x}}\right)

=tan^{-1} \left(\dfrac{\dfrac{x(1+x)+1}{1+x}}{\dfrac{1+x-x}{1+x}}\right)

=tan^{-1} \dfrac{x+x^2+1}{1}

= tan-1(1+x+x2) R.H.S

Proved


(iii) tan^{-1} \dfrac{1}{x+y} + tan^{-1} \dfrac{1}{x^2+xy+1}=cot^-{1} x

Sol :

L.H.S=tan^{-1} \dfrac{1}{x+y}+tan^{-1} \dfrac{y}{x+xy+1}

=tan^{-1}  \left(\dfrac{\dfrac{1}{x+y}+\dfrac{y}{x^2+xy+1}}{1-\dfrac{1}{x+y}\times \dfrac{y}{x^2+xy+1}}\right)

=tan^{-1} \dfrac{\left(\dfrac{x^2+xy+1+xy+y^2}{(x+y)(x^2+xy+1)}\right)}{\left(\dfrac{(x+y)(x^2+xy+1)-y}{(x+y)(x^2+xy+1)}\right)}

=tan^{-1} \dfrac{x^2+xy+1+xy+y^2}{x^3+xy^2+x+yx^2+xy^2+y-y}

=tan^{-1} \dfrac{(x^2+y^2+2xy+1)}{(x^3+xy^2+2x^2y+x)}

=tan^{-1} \dfrac{(x^2+y^2+2xy+1)}{(x(x^2+y^2+2xy+1))}

=tan^{-1} \dfrac{1}{x}=cot^{-1} xR.H.S

proved


(iv) 2 cot-1 5+cot-1 7+2 cot-18=π/4

Sol :

L.H.S=2 cot-1 5 + cot-17+2 cot-18

=2 cot-1 5 +2 cot-18+ cot-17

=2(cot-1 5+cot-1 8)+cot-1 7

=\left(tan^{-1} \dfrac{1}{5}+tan^{-1} \dfrac{1}{8}\right)+tan^{-1} \dfrac{1}{7}

=2\left(tan^{-1} \dfrac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5}\times \frac{1}{8}}\right)+tan^{-1} \dfrac{1}{7}

=2 tan^{-1}\left(\dfrac{\dfrac{13}{40}}{\dfrac{39}{40}}\right)+tan^{-1} \dfrac{1}{7}

=2 tan^{-1} \dfrac{1}{3}+tan^{-1} \dfrac{1}{7}

=tan^{-1} \dfrac{2\times \dfrac{1}{3}}{1-\dfrac{1}{9}}+tan^{-1} \dfrac{1}{7}

=tan^{-1} \left(\dfrac{\frac{2}{9}}{\frac{8}{9}}\right)+tan^{-1} \dfrac{1}{7}

=tan^{-1} \left(\dfrac{3}{4}\right)+tan^{-1} \dfrac{1}{7}

=tan^{-1} \left(\dfrac{\dfrac{3}{4}+\dfrac{1}{7}}{1-\dfrac{3}{4}\times\dfrac{1}{7}}\right)

=tan^{-1} \dfrac{\dfrac{25}{28}}{\dfrac{25}{28}}=tan^{-1} 1

=tan^{-1} tan \dfrac{\pi}{4}= \dfrac{\pi}{4} R.H.S

Proved


(v) tan^{-1} \left(\dfrac{x}{8}\right)-tan^{-1}\left(\dfrac{x-y}{x+y}\right)=\dfrac{\pi}{4}

Sol :

L.H.S=tan^{-1} \left(\dfrac{x}{y}\right)-tan^{-1} \left(\dfrac{x-y}{x+y}\right)

=tan^{-1} \left\{\dfrac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\frac{x}{y} \times \frac{x-y}{x+y}}\right\}

=tan^{-1} \dfrac{\left(\dfrac{x(x+y)-y(x-y)}{y(x+y)}\right)}{\left(\dfrac{y(x+y)+x(x-y)}{y(x+y)}\right)}

=tan^{-1} \dfrac{[x^2+xy-yx+y^2]}{[yx+y^2+x^2-yx]}

=tan^{-1} \dfrac{[x^2+y^2]}{[x^2+y^2]}=tan^{-1} 1

=tan^{-1} tan \dfrac{\pi}{4}=\dfrac{\pi}{4} R.H.S


Question 7

Prove that :

(i) tan^{-1} \dfrac{a-b}{1+ab}+tan^{-1} \dfrac{b-c}{1+bc}+tan^{-1} \dfrac{c-a}{1+ca}=0 ab>-1 , bc>-1 , ca>-1

Sol :

L.H.S=tan^{-1} \dfrac{a-b}{1+ab}+tan^{-1} \dfrac{b-c}{1+bc}+tan^{-1} \dfrac{c-a}{1+ca}

=tan-1 a-tan-1 b+tan-1 b-tan-1 c+tan-1 c-tan-1 a

=tan-1 a-tan-1 a-tan-1 b+tan-1 b-tan-1 c+tan-1 c

=0 =R.H.S 

proved


(ii) tan^{-1} \dfrac{a^3-b^3}{1+a^3b^3}+tan^{-1} \dfrac{c^3-a^3}{1+b^3c^3}+tan^{-1} \dfrac{c^3-a^3}{1+c^3a^3}=0

Sol :

L.H.S=tan^{-1} \dfrac{a^3-b^3}{1+a^3b^3}+tan^{-1} \dfrac{c^3-a^3}{1+b^3c^3}+tan^{-1} \dfrac{c^3-a^3}{1+c^3a^3}

=tan-1 a3-tan-1 b3+tan-1 b3-tan-1 c3+tan-1 c3-tan-1 a3

=tan-1 a3-tan-1 a3-tan-1 b3+tan-1 b3-tan-1 c3+tan-1 c3

=0 R.H.S

proved


Question 8

(i) sin^{-1} \dfrac{3}{5}+sin^{-1} \dfrac{8}{17}=sin^{-1} \dfrac{77}{85}

Sol :

L.H.S=sin^{-1} \dfrac{3}{5}+sin^{-1} \dfrac{8}{17}

*** QuickLaTeX cannot compile formula:
=sin^{-1} \left[\dfracc{3}{5}.\sqrt{1-\dfrac{8^2}{17^2}}+\dfrac{8}{17}.\sqrt{1-\left(\dfrac{3}{5}\right)^2}\right]

*** Error message:
Undefined control sequence \dfracc.
leading text: $=sin^{-1} \left[\dfracc

*** QuickLaTeX cannot compile formula:
=sin^{-1} \left[\dfracc{3}{5} \times \sqrt{\dfrac{17^2-8^2}{17^2}}+\dfrac{8}{17} \times \sqrt{\dfrac{5^2-3^2}{5^2}}\right]

*** Error message:
Undefined control sequence \dfracc.
leading text: $=sin^{-1} \left[\dfracc

*** QuickLaTeX cannot compile formula:
=sin^{-1} \left[\dfracc{3}{5} \times \dfrac{\sqrt{289-64}}{17}+\dfrac{8}{17} \times \dfrac{\sqrt{25-9}}{5}\right]

*** Error message:
Undefined control sequence \dfracc.
leading text: $=sin^{-1} \left[\dfracc

*** QuickLaTeX cannot compile formula:
=sin^{-1} \left[\dfracc{3}{5} \times \dfrac{\sqrt{225}}{17}+\dfrac{8}{17} \times \dfrac{\sqrt{16}}{5}\right]

*** Error message:
Undefined control sequence \dfracc.
leading text: $=sin^{-1} \left[\dfracc

*** QuickLaTeX cannot compile formula:
=sin^{-1} \left[\dfracc{3}{5} \times \dfrac{15}{17}+\dfrac{8}{17} \times \dfrac{4}{5}\right]

*** Error message:
Undefined control sequence \dfracc.
leading text: $=sin^{-1} \left[\dfracc

=sin^{-1} \left[\dfrac{45}{85}+\dfrac{32}{85}\right]

=sin^{-1} \left[\dfrac{77}{85}\right]=R.H.S proved

 


(ii) cos^{-1} \dfrac{3}{5}+cos^{-1}\dfrac{12}{13}+cos^{-1} \dfrac{63}{65}=\dfrac{\pi}{2}

Sol :

L.H.S=cos^{-1} \dfrac{3}{5}+cos^{-1}\dfrac{12}{13}+cos^{-1} \dfrac{63}{65}

=cos^{-1} \left( \dfrac{3}{5} \times \dfrac{12}{13}-\sqrt{1-\left(\dfrac{3}{5}\right)^2} \times \sqrt{1-\dfrac{12}{13}}\right)+cos^{-1} \dfrac{63}{65}

=cos^{-1} \left( \dfrac{36}{65} -\sqrt{\dfrac{5^2-3^2}{5^2}} \times \sqrt{\dfrac{13^2-12^2}{13^2}}\right)+cos^{-1} \dfrac{63}{65}

=cos^{-1} \left( \dfrac{36}{65} -\dfrac{\sqrt{25-9}}{5} \times \dfrac{\sqrt{169-144}}{13^2}\right)+cos^{-1} \dfrac{63}{65}

=cos^{-1} \left( \dfrac{36}{65} -\dfrac{4}{5}\times \dfrac{5}{19}\right)+cos^{-1} \dfrac{63}{65}

=cos^{-1} \left(\dfrac{36}{65}-\dfrac{20}{65}\right)+cos^{-1} \dfrac{63}{65}

=cos^{-1} \dfrac{16}{65}+cos^{-1} \dfrac{63}{65}

=cos^{-1}\left(\dfrac{16}{65} \times \dfrac{63}{65} -\sqrt{1-\left(\dfrac{16}{65}\right)^2} \times \sqrt{1-\left(\dfrac{63}{65}\right)^2}\right)

=cos^{-1}\left(\dfrac{16}{65} \times \dfrac{63}{65} -\sqrt{\dfrac{65^2-16^2}{65^2}} \times \sqrt{\dfrac{65^2-63^2}{65^2}}\right)

=cos^{-1} \left(\dfrac{1008}{65\times 65}-\dfrac{\sqrt{(65+16)(65-16)}}{65} \times \dfrac{\sqrt{(65+63)(65-63)}}{65} \right)

*** QuickLaTeX cannot compile formula:
=cos^{-1} \left(\dfrac{1008}{65\times 65}-\dfrac{\sqrt{81\times 49}}{65} \times \dfrac{\sqrt{128\times 2}{65}} \right)

*** Error message:
Missing } inserted.
leading text: ...imes \dfrac{\sqrt{128\times 2}{65}} \right

=cos^{-1} \left(\dfrac{1008}{65\times 65}-\dfrac{9\times 7}{65} \times \dfrac{16}{65} \right)

=cos^{-1} \left(\dfrac{1008}{65\times 65}-\dfrac{1008}{65\times 65} \right)

=cos^{-1} (0)=cos^{-1} cos \dfrac{\pi}{2}=\dfrac{\pi}{2} R.H.S Proved

 


(iii) sin^{-1} \dfrac{8}{17}+sin^{-1} \dfrac{3}{5}=tan^{-1} \dfrac{77}{36}=cos^{-1} \dfrac{36}{85}

Sol :

L.H.S=sin^{-1} \dfrac{8}{17}+sin^{-1} \dfrac{3}{5}

=sin^{-1} \left(\dfrac{8}{17}\times \sqrt{1-\left(\dfrac{3}{5}\right)^2}+ \dfrac{3}{5}\times \sqrt{1-\left(\dfrac{8}{17}\right)^2} \right)

=sin^{-1} \left(\dfrac{8}{17}\times \sqrt{\dfrac{5^2-3^2}{5^2}}+ \dfrac{3}{5}\times \sqrt{\dfrac{17^2-8^2}{17^2}} \right)

=sin^{-1} \left(\dfrac{8}{17}\times \dfrac{4}{5}+ \dfrac{3}{5}\times \dfrac{15}{17} \right)

=sin^{-1} \left( \dfrac{32}{85} + \dfrac{45}{85}\right)=sin^{-1} \dfrac{77}{85}

Let sin^{-1} \dfrac{77}{85}=\thetathensin \theta=\dfrac{77}{85}tan \theta =\dfrac{77}{36}∴\theta = tan^{-1} \dfrac{77}{36}sin^{-1} \dfrac{77}{85}= tan^{-1} \dfrac{77}{36}

*** QuickLaTeX cannot compile formula:
+ R.H.S proved

 

<hr />

<strong>(iv) </strong>

*** Error message:
Missing $ inserted.

cos^{-1} \dfrac{12}{13}+sin^{-1} \dfrac{5}{13}=sin^{-1} \dfrac{56}{65}

*** QuickLaTeX cannot compile formula:
question may be incorrect

<hr />

<strong>(v) </strong>

*** Error message:
Missing $ inserted.

cos^{-1} \dfrac{4}{5}+cos^{-1} \dfrac{12}{13}=cos^{-1} \dfrac{33}{65}

*** QuickLaTeX cannot compile formula:
Sol :

L.H.S

*** Error message:
Missing $ inserted.

=cos^{-1} \dfrac{4}{5}+cos^{-1} \dfrac{12}{13}=cos^{-1} \left(\dfrac{4}{5} \times \dfrac{12}{13} – \sqrt{1-\dfrac{16}{25} \times \sqrt{1-\dfrac{144}{169}}\right)=cos^{-1} \left( \dfrac{48}{65}-\dfrac{3}{5} \times \dfrac{5}{13}\right)=cos^{-1} \left(\dfrac{48}{65}-\dfrac{15}{65}\right)=cos^{-1} \dfrac{33}{65}

*** QuickLaTeX cannot compile formula:
= R.H.S proved

 

<hr />

<strong>(vi) </strong>

*** Error message:
Missing $ inserted.

sin^{-1} \dfrac{3}{5}-sin^{-1} \dfrac{8}{17}=cos^{-1} \dfrac{84}{85}

*** QuickLaTeX cannot compile formula:
Sol :

L.H.S

*** Error message:
Missing $ inserted.

sin^{-1} \dfrac{3}{5}-sin^{-1} \dfrac{8}{17}=sin^{-1} \left(\dfrac{3}{5} \times \sqrt{1-\left(\dfrac{8}{17}\right)^2}-\dfrac{8}{17} \times \sqrt{1-\left(\dfrac{9}{5}\right)^2}\right)$

working


(vii)

(viii)

(ix)

(x)

(xi)

 


Question 9

 

 


Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

later


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