SSC CGL 2024 PYQ — Question 69

The correct answer is Option 2 i.e. $2+\sqrt3 \over2$.

${tan(x)\over sec(x)} = {1\over 2}$

⇒ ${{sin(x)\over cos(x)}\over {1\over cos(x)}} = {1\over 2}$

⇒ $sin(x) = {1\over 2} = {P\over H}$ 

By Pythagoras theorem

⇒ H² = B² + P²

⇒ 2² = B² + 1²

⇒ B² = 4 - 1

⇒ B² = 3

⇒ B = $\sqrt3$

so, cos(x) = $B\over H$ = $\sqrt3 \over 2$

Now, (sinx + cos x)² = sin²x + cos²x + 2sinx.cosx

⇒ 1 + 2 × $\frac 1 2$ × $\sqrt3 \over 2$ = 1 + $\sqrt3 \over 2$ = $2+\sqrt3 \over2$