SSC CHSL 2024 PYQ — Question 61

The correct answer is option 2 i.e. $5\sqrt{2}-4\over4$.

tan A = P/B = $\sqrt2$ - 1 

P =  $\sqrt2$ - 1 , B = 1

By using the Pythagoras theorem

H² = P² + B²

⇒ H² = ($\sqrt2$ - 1)² + 1²

⇒ H² = $\sqrt{4-2\sqrt2}$ 

sinAcosA + tanA = $\frac{\sqrt2-1}{\sqrt{4-2\sqrt2}}\times\frac{1}{\sqrt{4-2\sqrt2}}+\sqrt2-1$

= $\frac{\sqrt2-1}{ 4-2\sqrt2}+\sqrt2-1$

= $\frac{\sqrt2-1}{ 2\sqrt2(\sqrt2-1)}+\sqrt2-1$

= $\frac{1}{2\sqrt2}+\sqrt2-1$

= $\frac{1}{2\sqrt2}\times\frac{\sqrt2}{\sqrt2}+\sqrt2-1$

= $\frac{\sqrt2+4\sqrt2-4}{4}$

= $5\sqrt{2}-4\over4$