Question 1. Find the value of [Tex]{\cos }^{-1}(\cos \frac {13\pi} {6})[/Tex]
Solution:
We know that [Tex]\cos^{-1} (\cos x)=x [/Tex]
Here, [Tex]\frac {13\pi} {6} \notin [0,\pi].[/Tex]
Now, [Tex]{\cos }^{-1}(\cos \frac {13\pi} {6}) [/Tex] can be written as :
[Tex]{\cos }^{-1}(\cos \frac {13\pi} {6})={\cos }^{-1}[\cos( 2\pi+\frac {\pi} {6})] [/Tex], where [Tex]\frac{\pi} {6} \in [0,\pi].[/Tex]
Hence, the value of [Tex]{\cos }^{-1}(\cos \frac {13\pi} {6}) [/Tex] = π/6
Question 2. Find the value of [Tex]\tan^{-1}(\tan \frac {7\pi}{6})[/Tex]
Solution:
We know that [Tex]\tan^{-1} (\tan x)=x[/Tex]
Here, [Tex]\frac {7\pi}{6} \notin(\frac{-\pi}{2},\frac{\pi}{2})[/Tex]
Now, [Tex]\tan^{-1}(\tan \frac {7\pi}{6}) [/Tex] can be written as:
[Tex]\tan^{-1}(\tan \frac {7\pi}{6})=\tan^{-1}[\tan( 2\pi -\frac {5\pi}{6})] [/Tex] [Tex]-[\tan(2\pi-x)=-\tan x][/Tex]
[Tex]\tan^{-1}[-\tan(\frac {5\pi}{6}) ]=\tan^{-1}[\tan(-\frac {5\pi}{6})][/Tex]
[Tex]=\tan^{-1}[\tan(\pi-\frac {5\pi}{6})]=\tan^{-1}[\tan( \frac {\pi}{6})], [/Tex] where [Tex]\frac{\pi}{6} \in (\frac{-\pi}{2},\frac{\pi}{2})[/Tex]
Hence, the value of [Tex]\tan^{-1}(\tan\frac{7\pi}{6}) [/Tex] = π/6
Question 3. Prove [Tex]2\sin^{-1} \frac{3}{5}=\tan^{-1} \frac{24}{7}[/Tex]
Solution:
Let [Tex]\sin^{-1} \frac{3}{5}=x [/Tex] -(1)
sin x = 3/5
So,[Tex] \cos x = \sqrt{1-(\frac{3}{5})^2 } [/Tex]= 4/5
tan x = 3/4
Hence, [Tex]x=\tan^{-1} \frac{3}{4} [/Tex]
Now put the value of x from eq(1), we get
[Tex]\sin^{-1} \frac{3}{5}=\tan^{-1} \frac{3}{4}[/Tex]
Now, we have
L.H.S [Tex]= 2 \sin^{-1} \frac{3}{5}=2 \tan^{-1} \frac{3}{4}[/Tex]
= [Tex]\tan^{-1}(\frac{2 \times \frac{3}{4}}{1-(\frac{3}{4})^{2}}) [/Tex] -[Tex][2\tan^{-1} x=\tan^{-1} \frac{2x}{1-x^2}][/Tex]
[Tex]= \tan^{-1}(\frac{ \frac{3}{2}}{\frac{16-9}{16}})=\tan^{-1} (\frac{3}{2} \times \frac{16}{7})[/Tex]
[Tex]=\tan^{-1} \frac{24}{7} [/Tex]
Hence, proved.
Question 4. Prove [Tex]\sin^{-1} \frac{8}{17}+\sin^{-1} \frac{3}{5}=\tan^{-1} \frac{77}{36}[/Tex]
Solution:
Let [Tex]\sin^{-1} \frac{8}{17}=x [/Tex]
Then sin x = 8/17
cos x =[Tex] \sqrt{1-(\frac{8}{17})^2}=\sqrt \frac{225}{289} [/Tex] = 15/17
Therefore, [Tex]\tan x=\frac{8}{15}\implies x=\tan^{-1}\frac{8}{15}[/Tex]
[Tex]\sin^{-1} \frac{8}{17}=\tan^{-1} \frac{8}{17} [/Tex] -(1)
Now, let [Tex]\sin^{-1} \frac{3}{5}=y [/Tex]
Then, sin y = 3/5
[Tex]\cos y=\sqrt{1- (\frac{3}{5})^2}=\sqrt{ (\frac{16}{25})} [/Tex] = 4/5
[Tex]\therefore \tan y =\frac{3}{4} \implies y=\tan^{-1} \frac{3}{4}[/Tex]
[Tex]\therefore \sin^{-1} \frac{3}{5}=\tan^{-1} \frac{3}{4} [/Tex] -(2)
Now, we have:
L.H.S.[Tex]=\sin^{-1} \frac{8}{17}+\sin^{-1} \frac{3}{5}[/Tex]
From equation(1) and (2), we get
= [Tex]\tan^{-1} \frac{8}{15}+\tan^{-1} \frac{3}{4} [/Tex]
= [Tex]\tan^{-1} \frac{{\frac{8}{15}+ \frac{3}{4}} }{1-{\frac{8}{15}\times \frac{3}{4}}}[/Tex]
= [Tex]\tan^{-1}(\frac{32+45}{60-24}) [/Tex] -[Tex][\tan^{-1} x + \tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}] [/Tex]
= [Tex]\tan^{-1} \frac{77}{36} [/Tex]
Hence proved
Question 5. Prove [Tex]\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}[/Tex]
Solution:
Let [Tex]\cos^{-1}\frac{4}{5}= x [/Tex]
Then, cos x = 4/5
[Tex]\sin x = \sqrt {1- (\frac{4}{5})^{2}} [/Tex] = 3/5
[Tex]\therefore \tan x =\frac{3}{4} \implies x=\tan^{-1} \frac{3}{4}[/Tex]
[Tex]\therefore \cos^{-1} \frac{4}{5}=\tan^{-1} \frac{3}{4} [/Tex] -(1)
Now let [Tex]\cos^{-1} \frac{12}{13}=x [/Tex]
Then, cos y = 3/4
[Tex]\sin^{-1} y=\frac{5}{13}[/Tex]
[Tex]\therefore\tan y= \frac{5}{12} \implies y=\tan^{-1} \frac{5}{12}[/Tex]
[Tex]\therefore \cos ^{-1} \frac{12}{13}=\tan^{-1} \frac{5}{12} [/Tex] -(2)
Let [Tex]\cos^{-1} \frac{33}{65}=z [/Tex]
Then, cos z = 33/65
sin z = 56/65
[Tex]\therefore \tan z = \frac{56}{65} \implies z= tan^{-1}\frac{56}{33}[/Tex]
[Tex]\therefore \cos^{-1} \frac{33}{65}= \tan^{-1} \frac{56}{33} [/Tex] -(3)
Now, we will prove that :
L.H.S. [Tex]=\cos^{-1} \frac{3}{5}\cos^{-1} \frac{12}{13}[/Tex]
From equation (1) and equation (2)
= [Tex]\tan^{-1} \frac{3}{4}+\tan^{-1} \frac{5}{12} [/Tex]
= [Tex]\tan^{-1} \frac{{\frac{3}{4}+ \frac{5}{12}} }{1-{\frac{3}{4}\times \frac{5}{12}}} [/Tex] -[Tex][\tan^{-1} x +\tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}][/Tex]
= [Tex]\tan^{-1} \frac{36+20}{48-15}[/Tex]
= [Tex]\tan^{-1} \frac{56}{33}[/Tex]
Using equation(3)
= [Tex]\tan^{-1} \frac{56}{33} [/Tex]
Hence proved
Question 6. Prove [Tex]\cos^{-1} \frac{12}{13}+\sin^{-1} \frac{3}{5}=\sin^{-1} \frac{56}{65}[/Tex]
Solution:
Let [Tex]\sin^{-1} \frac{3}{5}=x [/Tex]
Then, sin x = 3/5
[Tex]\cos x =\sqrt{1- (\frac{3}{5})^{2}}=\sqrt \frac{16}{25} [/Tex] = 4/5
[Tex]\therefore \tan x = \frac{3}{4} \implies x= \tan^{-1} \frac{3}{4}[/Tex]
[Tex]\therefore \sin^{-1} \frac{3}{5}= \tan^{-1} \frac{3}{4} [/Tex] -(1)
Now, let [Tex]\cos^{-1} \frac{12}{13}=y [/Tex]
Then, cos y = 12/13 and sin y = 5/13
[Tex]\therefore \tan y = \frac{5}{12} \implies y= \tan^{-1} \frac{5}{12}[/Tex]
[Tex]\therefore \cos^{-1} \frac{12}{13}= \tan^{-1} \frac{5}{12} [/Tex] -(2)
Let [Tex]\sin^{-1} \frac{56}{65}=z[/Tex]
Then, sin z = 56/65 and cos z = 33/65
[Tex]\therefore \tan z = \frac{56}{33} \implies z=\tan ^{-1} \frac{56}{33}[/Tex]
[Tex]\therefore \sin^{-1} \frac{56}{65}= \tan^{-1} \frac{56}{33} [/Tex] -(3)
Now, we have:
L.H.S.=[Tex]\cos^{-1} \frac{12}{13}+ \sin^{-1} \frac{3}{5}[/Tex]
From equation(1) and equation(2)
=[Tex]\tan^{-1} \frac{5}{12}+\tan^{-1} \frac{3}{4} [/Tex]
= [Tex]\tan^{-1} \frac{{\frac{5}{12}+ \frac{3}{4}} }{1-{\frac{5}{12}\times \frac{3}{4}}} [/Tex] -[Tex][\tan^{-1} x +\tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}] [/Tex]
= [Tex]\tan^{-1} \frac{20+36}{48-15}[/Tex]
= [Tex]\tan^{-1} \frac{56}{33}[/Tex]
From equation (3)
= [Tex]\sin^{-1} \frac{56}{65} [/Tex]
Hence proved
Question 7. Prove [Tex]\tan^{-1} \frac{63}{16}= \sin^{-1} \frac{5}{13}+\cos^{-1} \frac{3}{5}[/Tex]
Solution:
Let [Tex]\sin^{-1} \frac{5}{13}=x [/Tex]
Then, sin x = 5/13 and cos x = 12/13.
[Tex]\tan^{-1} \frac{7+5}{35-1}+\tan^{-1} \frac{8+3}{24-1}[/Tex]
[Tex]\therefore \tan x= \frac{5}{12} \to x= \tan^{-1} \frac{5}{12}[/Tex]
[Tex]\therefore \sin^{-1} \frac{5}{13}= \tan^{-1} \frac{5}{12} [/Tex] -(1)
Let [Tex]\cos^{-1} \frac{3}{5}=y [/Tex]
Then, cos y = 3/5 and sin y = 4/5
[Tex]\therefore \tan y= \frac{4}{3} \implies y= \tan^{-1}\frac{4}{3}[/Tex]
[Tex]\therefore \cos ^{-1}\frac{3}{5}=\tan^{-1} \frac{4}{3} [/Tex] -(2)
From equation(1) and (2), we have
R.H.S.[Tex]=\sin^{-1} \frac{5}{13}+\cos^{-1} \frac{3}{5}[/Tex]
=[Tex]\tan^{-1} \frac{5}{12}+\tan^{-1} \frac{4}{3}[/Tex]
= [Tex]\tan^{-1} \frac{{\frac{5}{12}+ \frac{4}{3}} }{1-{\frac{5}{12}\times \frac{4}{3}}} [/Tex] -[Tex][\tan^{-1} x +\tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}][/Tex]
=[Tex]\tan^{-1} \frac{15+48}{36-20}[/Tex]
=[Tex]\tan^{-1} \frac{63}{16}[/Tex]
L.H.S = R.H.S
Hence proved
Question 8. Prove [Tex]\tan^{-1} \frac{1}{5}+\tan^{-1} \frac{1}{7}\tan^{-1} \frac{1}{3}+\tan^{-1} \frac{1}{8}=\frac{\pi}{4}[/Tex]
Solution:
L.H.S.[Tex]=\tan^{-1} \frac{1}{5}+\tan^{-1} \frac{1}{7}\tan^{-1} \frac{1}{3}+\tan^{-1} \frac{1}{8}[/Tex]
= [Tex]\tan^{-1} \frac{{\frac{1}{5}+ \frac{1}{7}} }{1-{\frac{1}{5}\times \frac{1}{7}}} +\tan^{-1} \frac{{\frac{1}{3}+ \frac{1}{8}} }{1-{\frac{1}{3}\times \frac{1}{8}}} [/Tex] -[Tex][\tan^{-1} x +\tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}][/Tex]
= [Tex]\tan^{-1} \frac{7+5}{35-1}+\tan^{-1} \frac{8+3}{24-1}[/Tex]
= [Tex]\tan^{-1} \frac{12}{34}+\tan^{-1} \frac{11}{23}[/Tex]
= [Tex]\tan^{-1} \frac{6}{17}+\tan^{-1} \frac{11}{23}[/Tex]
= [Tex]\tan^{-1} \frac{{\frac{6}{17}+ \frac{11}{23}} }{1-{\frac{6}{17}\times \frac{11}{23}}} [/Tex]
= [Tex]\tan^{-1} \frac{138 + 187}{391-66}[/Tex]
= [Tex]\tan^{-1} \frac{325}{325}=\tan^{-1} 1[/Tex]
= π/4
L.H.S = R.H.S
Hence proved
Question 9. Prove [Tex]\tan^{-1} \sqrt x= \frac{1}{2} \cos^{-1} (\frac{1-x}{1+x}),x\in[0,1][/Tex]
Solution:
Let x = tan2θ
Then,[Tex]\sqrt x=\tan \theta \implies\theta=\tan^{-1} \sqrt x.[/Tex]
[Tex]\therefore \frac{1-x}{1+x}+\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}=\cos 2\theta[/Tex]
Now, we have
R.H.S = [Tex]\frac{1}{2} \cos ^{-1}(\frac{1-x}{1+x})= \frac{1}{2} \cos ^{-1} (\cos 2 \theta)=\frac{1}{2} \times 2 \theta=\theta=\tan^{-1}\sqrt x[/Tex]
L.H.S = R.H.S
Hence proved
Question 10. Prove [Tex]\cot^{-1} (\frac{\sqrt ({1+ \sin x}) + \sqrt ({1-\sin x})} {\sqrt ({1+ \sin x})-\sqrt ({1- \sin x})})=\frac{x}{2},x\in(0,\frac{\pi}{4})[/Tex]
Solution:
Consider[Tex] (\frac{\sqrt ({1+ \sin x}) + \sqrt ({1-\sin x})} {\sqrt ({1+ \sin x})-\sqrt ({1- \sin x})})[/Tex]
By rationalizing
=[Tex] \frac{(\sqrt ({1+ \sin x}) + \sqrt ({1-\sin x}))^{2}}{(\sqrt{ ({1+ \sin x})}-\sqrt({1- \sin x}))^{2}} [/Tex]
=[Tex] \frac{( {1+ \sin x)} + {(1-\sin x)} + 2 \sqrt{(1+\sin x)(1-\sin x)}}{{ {1+ \sin x}}-{1+\sin x}}[/Tex]
=[Tex]\frac{2(1+\sqrt{1-\sin^{2}})}{2\sin x}=\frac{1+\cos x}{\sin x}=\frac{2\cos ^{2}\frac{x}{2}}{2\sin \frac{x}{2}\cos\frac{x}{2}} [/Tex]
= [Tex]\cot \frac{x}{2}[/Tex]
L.H.S = [Tex]\cot^{-1} (\frac{\sqrt ({1+ \sin x}) + \sqrt ({1-\sin x})} {\sqrt ({1+ \sin x})-\sqrt ({1- \sin x})})={\cot}^{-1}({\cot( \frac x 2)}) [/Tex]= x/2
L.H.S = R.H.S
Hence proved
Share your thoughts in the comments
Please Login to comment...