Find the Value of sec x + tan x

Question:

\[ \sin x = \frac{12}{13} \]

and \(x\) lies in the second quadrant, find the value of:

\[ \sec x + \tan x \]

Solution

Given,

\[ \sin x = \frac{12}{13} \]

Since \(x\) lies in Quadrant II, sine is positive while cosine and tangent are negative.

Using:

\[ \sin x = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \]

Take,

\[ \text{Perpendicular} = 12, \quad \text{Hypotenuse} = 13 \]

Using Pythagoras theorem:

\[ \text{Base} = \sqrt{13^2 – 12^2} \]

\[ = \sqrt{169 – 144} \]

\[ = \sqrt{25} = 5 \]

In Quadrant II, base is negative.

Therefore,

\[ \cos x = -\frac{5}{13} \]

\[ \sec x = \frac{1}{\cos x} = -\frac{13}{5} \]

\[ \tan x = \frac{\sin x}{\cos x} = \frac{\frac{12}{13}}{-\frac{5}{13}} = -\frac{12}{5} \]

Now,

\[ \sec x + \tan x = -\frac{13}{5} – \frac{12}{5} \]

\[ = -\frac{25}{5} \]

\[ = -5 \]

\[ \boxed{\sec x + \tan x = -5} \]

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