SSC CGL 2022 PYQ — Question 64

The correct answer is option 4 i.e.  $\frac{3}{\sqrt5}$

Formula used:

When a quadratic equation is given i.e. ax² + bx + c = 0

The roots of the equation are given as:

$x = \dfrac{-b \pm \sqrt{b^2- 4ac}}{2a}$        ----- (1)

Calculations:

⇒ 6tanA(tanA + 1) = 5 - tanA

⇒ 6tan²A + 6tanA = 5 - tanA

⇒ 6tan²A + 7tanA - 5 = 0

Using equation (1), we get

tanA = 1/2, -5/3

We will take the value of tanA = 1/2 = perpendicular/base

Now using Pythagoras theorem

Hypotenuse² = (√4 + 1)

Hypotenuse = √5

cosA = 2/√5 and sinA = 1/√5

(sinA + cosA) = 2/√5 + 1/√5 = 3/√5