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  • If  <img alt="\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1" src="https://learn.careers360.com/latex-image/?%5Csin%20%5Cleft%28%5Csin%20%5E%7B-1%7D%20%5Cfrac%7B1%7D%7B5%7D&plus;%

If  \sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1 then x is equal to

Option: 1 1

Option: 2 0

Option: 3 0.8

Option: 4 0.2

Answers (1)

best_answer

 

 

Different Inverse trigonometric function relating with each other -

Different Inverse trigonometric function relating with each other

We can write the different trigonometric function in the form of other trigonometric functions

Let sin-1 x = Θ. Then sin Θ = x or sin Θ = x/1 

So, we have the right-angled triangle

 

From the figure

 

\\\mathrm{\cos\theta=\frac{\sqrt{1-x^2}}{1}=\sqrt{1-x^2}}\\\\\mathrm{\therefore \quad\quad\theta=\sin^{-1}x=\cos^{-1}\sqrt{1-x^2}}\\\\\text{Also, we have}\;\;\theta=\tan^{-1}\frac{x}{\sqrt{1-x^2}}\\\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=cot^{-1}\frac{\sqrt{1-x^2}}{x}}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=sec^{-1}\frac{1}{\sqrt{1-x^2}}}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=csc^{-1}\frac{1}{x}}


 

Now for Θ = cos-1 x, x > 0 or cos Θ = x we have the following right angled triangle

 

 

From the figure

 

\\\mathrm{\sin\theta=\frac{\sqrt{1-x^2}}{1}=\sqrt{1-x^2}}\\\\\mathrm{\therefore \quad\quad\theta=\cos^{-1}x=\sin^{-1}\sqrt{1-x^2}}\\\\\text{Also, we have}\;\;\theta=\tan^{-1}\frac{\sqrt{1-x^2}}{x}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=cot^{-1}\frac{x}{\sqrt{1-x^2}}}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=sec^{-1}\frac{1}{x}}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=csc^{-1}\frac{1}{\sqrt{1-x^2}}} 



 

Now for Θ = tan-1 x, x > 0 or tan Θ = x we have the following right angled triangle

 

 

From the figure

 

\\\mathrm{\sin\theta=\frac{x}{\sqrt{1+x^2}}}\\\\\mathrm{ \quad\quad\theta=\tan^{-1}x=\sin^{-1}\frac{x}{\sqrt{1+x^2}}}\\\\\text{Also, we have}\;\;\theta=\cos^{-1}\frac{1}{\sqrt{1+x^2}}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=cot^{-1}\frac{1}{x}}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=sec^{-1}\sqrt{1+x^2}}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=csc^{-1}\frac{\sqrt{1+x^2}}{x}}

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\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1\rightarrow \sin ^{-1} \frac{1}{5}+\cos ^{-1} x=\frac{\pi}{2}\\ \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} \frac{1}{5}\\ \cos ^{-1} x=\cos ^{-1} \frac{1}{5}\\ x=\frac{1}{5}

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seema garhwal

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