Given:
Sin x + cos x = 0 and x lies in fourth quadrant.
Sin x = -cos x
\(\dfrac{ Sin x}{cos x}\) = -1
So, tan x = -1 (since, tan x = sin x/cos x)
We know that, in fourth quadrant, cos x and sec x are positive and all other ratios are negative.
By using the formulas,
Sec x = \( \sqrt{(1 + tan^2 x)}\)
Cos x =\( \dfrac{ 1}{sec x}\)
Sin x = \( - \sqrt{(1 -cos^2 x)}\)
Now,
Sec x = \( \sqrt{(1 + tan^2 x)}\)
= \( \sqrt{(1 + (-1)^2)}\)
= \( \sqrt{2}\)
Cos x = \( \dfrac{ 1}{sec x}\)
= \( \dfrac{1}{\sqrt{2}}\)
Sin x = \( – \sqrt{(1 – cos^2 x)}\)
= \( \sqrt{-\dfrac{1}{2}}\)
= \( - \dfrac{1}{\sqrt{2}}\)
sin x = \( - \dfrac{1}{\sqrt{2}}\) and cos x = \( \dfrac{1}{\sqrt{2}}\)
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