SSC CGL 7th March 2020 Shift-2 Question 73

Question 73

If $$\frac{\sin A + \cos A}{\cos A} = \frac{17}{12}$$, then the value of $$\frac{1 - \cos A}{\sin A}$$ is:

Solution

$$\frac{\sin A + \cos A}{\cos A} = \frac{17}{12}$$

$$\frac{\cos A(\frac{\sin A}{\cos A} + 1)}{\cos A} = \frac{17}{12}$$

tan A + 1 = $$\frac{17}{12}$$

tan A =  $$\frac{17}{12}$$ - 1 = $$\frac{5}{12}$$

By triplet 5-12-13,

Perpendicular = 5

base = 12

hypotenuse = 13 

so,

Sin A = perpendicular/hypotenuse = 5/13

cos A = base/hypotenuse = 12/13

Now,

$$\frac{1 - \cos A}{\sin A}$$

= $$\frac{1 - \frac{12}{13}}{\frac{5}{13}}$$ = 1/5



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