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

#### Multiple Choice Question (MCQ)

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

1. -\dfrac{4}{3}

2. ×

-\dfrac{4}{3} or -\dfrac{3}{4}

3. ×

-\dfrac{3}{4}

4. ×

\dfrac{3}{4} or -\dfrac{4}{3}

Collected in the board: Trigonometry

Steven Zheng posted 5 months ago

1. 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}
(1)

Apply double angle identity for sine function,

\sin 2x = - \dfrac{24}{25}

\pi <2x<2\pi

\dfrac{\pi}{2} < x< \pi
(2)
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,

\tan = \dfrac{\sin x}{\cos x} = -\dfrac{4}{3}

A is the choice

Steven Zheng posted 5 months ago

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