If $$sin x = \dfrac{12}{13}$$ and lies in the second quadrant, find the value of sec x + tan x.

Asked by Sakshi | 1 year ago |  77

##### Solution :-

Given:

Sin x = $$\dfrac{12}{13}$$ and x lies in the second quadrant.

We know, in second quadrant, sin x and cosec x are positive and all other ratios are negative.

By using the formulas,

Cos x = $$\sqrt{(1-sin^2 x)}$$

$$\dfrac{-5}{13}$$

We know,

tan x =$$\dfrac{sin x}{cos x}$$

sec x = $$\dfrac{1}{cos x}$$

Now,

tan x =$$\dfrac{-12}{5}$$

sec x = $$\dfrac{-13}{5}$$

Sec x + tan x = $$\dfrac{-13}{5}+$$ +$$\dfrac{-12}{5}$$

$$\dfrac{-25}{5}$$

= -5

Sec x + tan x = -5

Answered by Sakshi | 1 year ago

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