If $$\sin x+\cos x = \dfrac{1}{5}$$, and $$0

Question

If $$\sin x+\cos x = \dfrac{1}{5}$$, and $$0 < x <\pi$$, then the value of $$	an x $$

Uniteasy Admin · Accepted Answer

Square the both sides of the equation of
$$\sin x+\cos x = \dfrac{1}{5}$$
We get,
$$\sin^2 x+\cos^2 x+2\sin x\cos x = \dfrac{1}{25} $$
Apply Pythagorean Identity
$$2\sin x\cos x = - \dfrac{24}{25}  $$n
Apply double angle identity for sine function,
$$\sin 2x = - \dfrac{24}{25}$$
$$\pi <2x<2\pi$$
$$\dfrac{\pi}{2}  < x< \pi$$n
$$1- 2\sin x\cos x = 1-( - \dfrac{24}{25}) = \dfrac{49}{25}  $$

$$(\sin x-\cos x)^2 =  \dfrac{49}{25}$$
$$\sin x-\cos x = \dfrac{7}{5} $$ (because $$ \dfrac{\pi}{2}  < x< \pi$$) 
Solve the  equation system 
$$\sin x-\cos x = \dfrac{7}{5} $$
$$\sin x+\cos x = \dfrac{1}{5}$$
Wr get
$$\sin x = \dfrac{4}{5}, \cos x = -\dfrac{3}{5}  $$
Therefore,
$$	an = \dfrac{\sin x}{\cos x}  = -\dfrac{4}{3}  $$

A is the choice

Multiple Choice Question (MCQ)

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