Question

If ${\cos }^{-1}x={\tan }^{-1}x,$ then

• A. $x^2=\frac{\sqrt{5}-1}{2}$
• B. there are two such values of $x$ in $[-1,1]$
• C. $\sin \left({\cos }^{-1}x\right)=\frac{\sqrt{5}-1}{2}$
• D. $\tan \left({\cos }^{-1}x\right)=\frac{\sqrt{5}-1}{2}$

Answer 1

Answer: (A), (C)

$\sin \left({\cos }^{-1}x\right)=\frac{\sqrt{5}-1}{2}$

${\cos }^{-1}x={\tan }^{-1}x\cdots \left(1\right)$
Clearly, the equation is defined for $x\in [-1,1]$
Considering the range of ${\cos }^{-1}x$ and ${\tan }^{-1}x$ in $[-1,1],$ we can conclude that the equation is satisfied only for some value of LHS and RHS $\in [0,\frac{\pi }{4}]$
This means $x\in [0,1]$

Now, ${\cos }^{-1}x={\tan }^{-1}x$
$\Rightarrow {\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)={\tan }^{-1}x$
$\Rightarrow \frac{\sqrt{1-x^2}}{x}=x$
$\Rightarrow \sqrt{1-x^2}=x^2$
$\Rightarrow 1-x^2=(x^2{)}^2$
$\Rightarrow (x^2{)}^2+x^2-1=0$
$\Rightarrow x^2=\frac{-1+\sqrt{5}}{2}$
Since $x\in [0,1],$ there is only one such $x.$

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

$\tan \left({\cos }^{-1}x\right)=\tan \left({\tan }^{-1}x\right)=x\left[\text{From}\right(1\left)\right]$