Question

Which of the following pairs of function/functions has same graph?

• A. $y=\tan \left({\cos }^{-1}x\right);y=\frac{\sqrt{1-x^2}}{x}$
• B. $y=\tan \left({\cot }^{-1}x\right);y=\frac{1}{x}$
  
• C. $y=\sin \left({\tan }^{-1}x\right);y=\frac{x}{\sqrt{1+x^2}}$
• D. $y=\cos \left({\tan }^{-1}x\right);y=\sin \left({\cot }^{-1}x\right)$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $y=\tan \left({\cos }^{-1}x\right);y=\frac{\sqrt{1-x^2}}{x}$
B $y=\tan \left({\cot }^{-1}x\right);y=\frac{1}{x}$

C $y=\sin \left({\tan }^{-1}x\right);y=\frac{x}{\sqrt{1+x^2}}$
D $y=\cos \left({\tan }^{-1}x\right);y=\sin \left({\cot }^{-1}x\right)$

$y=\tan \left({\cos }^{-1}x\right);y=\frac{\sqrt{1-x^2}}{x}$
Domain of the functions
$D_1=[-1,1]-\left\{0\right\}{D}_2=[-1,1]-\left\{0\right\}$
Now, we know that,
$y=\tan \left({\cos }^{-1}x\right)\Rightarrow y=\tan \left({\tan }^{-1}\frac{\sqrt{1-x^2}}{x}\right)\Rightarrow y=\frac{\sqrt{1-x^2}}{x}$
Therefore the function indentical, hence they will have same graph.

$y=\tan \left({\cot }^{-1}x\right);y=\frac{1}{x}$
$D_1=R-\left\{0\right\}{D}_2=R-\left\{0\right\}$
We know that,
$y=\tan \left({\cot }^{-1}x\right)\Rightarrow y=\tan \left({\tan }^{-1}\frac{1}{x}\right)=\frac{1}{x}$
Therefore the function indentical, hence they will have same graph.

$y=\sin \left({\tan }^{-1}x\right);y=\frac{x}{\sqrt{1+x^2}}$
Domain of both the function is $R$ and
$y=\sin \left({\tan }^{-1}x\right)\Rightarrow y=\sin \left({\sin }^{-1}\frac{x}{\sqrt{1+x^2}}\right)\Rightarrow y=\frac{x}{\sqrt{1+x^2}}$
Therefore the function indentical, hence they will have same graph.

$y=\cos \left({\tan }^{-1}x\right);y=\sin \left({\cot }^{-1}x\right)$
Domain of both the function is $R$ and
$y=\cos \left({\tan }^{-1}x\right)\Rightarrow y=\cos \left({\cos }^{-1}\frac{1}{\sqrt{1+x^2}}\right)\Rightarrow y=\frac{1}{\sqrt{1+x^2}}$
$y=\sin \left({\cot }^{-1}x\right)\Rightarrow y=\sin \left(\frac{1}{\sqrt{1+x^2}}\right)\Rightarrow y=\frac{1}{\sqrt{1+x^2}}$
Therefore the function indentical, hence they will have same graph.