SSC CHSL 2024 PYQ — Question 67

The correct answer is option 1 i.e. $sin^{-1} ({304\over 425})$.

$cos^{-1} ({7\over 25}) - cos^{-1} ({15\over 17})$

cos^-1x - cos^-1y = cos^-1(xy + $\sqrt{1-x^2}.\sqrt{1-y^2}$)

cos^-1(x/y) = sin^-1$\frac{\sqrt{y^2-x^2}}{y}$

⇒ $cos^{-1} ({7\over 25}) - cos^{-1} ({15\over 17})$ =  cos^-1($\frac{7}{25}\times\frac{15}{17}+\sqrt{1-(\frac{7}{25})^2}\sqrt{1-(\frac{15}{17})^2}$)

⇒ $cos^{-1} ({7\over 25}) - cos^{-1} ({15\over 17})$ =  cos^-1($\frac{105}{425}+\sqrt{\frac{576}{625}} \sqrt{\frac{64}{289}}$)

⇒ $cos^{-1} ({7\over 25}) - cos^{-1} ({15\over 17})$ =  cos^-1($\frac{105}{425}+\frac{192}{425}$)

⇒ $cos^{-1} ({7\over 25}) - cos^{-1} ({15\over 17})$ =  cos^-1($\frac{297}{425}$)

 cos^-1($\frac{297}{425}$) =  sin^-1$\frac{\sqrt{425^2-297^2}}{425}$ = $sin^{-1} ({304\over 425})$