Question

If $a^2=26/51$, then $b^2$ is equal to

• A. 1/25
• B. 24/25
• C. 25/26
• D. 50/51

Answer 1

Answer: (B)

24/25

$a=costa{n}^{-1}sinco{t}^{-1}x=costa{n}^{-1}sin{sin}^{-1}\left(\frac{\frac{1}{x}}{\sqrt{1+{\left(\frac{1}{x}\right)}^2}}\right)=costa{n}^{-1}sin{sin}^{-1}\left(\frac{1}{\sqrt{x^2+1}}\right)=costa{n}^{-1}\left(\frac{1}{\sqrt{x^2+1}}\right)=cos{cos}^{-1}\left(\frac{1}{\sqrt{1+{\left(\frac{1}{\sqrt{x^2+1}}\right)}^2}}\right)=cos{cos}^{-1}\left(\sqrt{\frac{x^2+1}{x^2+2}}\right)=\sqrt{\frac{x^2+1}{x^2+2}}$
For $a^2=\frac{26}{51}\Rightarrow \frac{x^2+1}{x^2+2}=\frac{26}{51}\Rightarrow x=\pm \frac{1}{5}$
Now
$b=cos(2{cos}^{-1}x+{sin}^{-1}x)=cos({cos}^{-1}x+\frac{\pi }{2})=-sin{cos}^{-1}x=-sin{sin}^{-1}\left(\sqrt{1-x^2}\right)=-\sqrt{1-x^2}\Rightarrow b^2=1-x^2=1-\frac{1}{25}=\frac{24}{25}$