Simplify $\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}$

Question

Which one is equivalent to: $$\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}$$
1)$\dfrac{\cos(x)}{1+\sin(x)}$
2)$\dfrac{\cos(x)}{1-\sin(x)}$
3)$\dfrac{1-\sin(x)}{\cos(x)}$
4)$\dfrac{1+\sin(x)}{\cos(x)}$

Of course we can find the correct choice by testing each of them,but I'm looking for an analytical solution assuming we don't know the final expression. I tried the following techniques,all failed: multiplying by $\frac{\sin(x)}{\sin(x)}$ , multiplying by $\frac{\cos(x)}{\cos(x)}$ , using $\sin^2(x)+\cos^2(x)=1$

https://math.stackexchange.com/questions/426981/need-help-in-proving-that-frac-sin-theta-cos-theta-1-sin-theta-cos — lab bhattacharjee, Nov 18 '17 at 18:56
HINT: Put $x=0$ to eliminate 1) and 2). Next put $ x= \pi/4$ only 3) remains. — Narasimham, Nov 19 '17 at 03:42
1 and 2 are equals and they give the result of your expresion see my answers belows — Guy Fsone, Nov 19 '17 at 17:10

score 1 · Answer 1 · answered Nov 18 '17 at 18:50

1

$$\cos2y=1-2\sin^2y=2\cos^2y-1,\sin2y=2\sin y\cos y$$

$$\dfrac{1-\sin2y+\cos2y}{1+\sin2y+\cos2y}=\dfrac{2\cos^2y-2\sin y\cos y}{2\cos^2y+2\sin y\cos y}=\dfrac{\cos y-\sin y}{\cos y+\sin y}$$

$$(i)=\dfrac{(\cos y-\sin y)^2}{\cos^2y-\sin^2y}=\dfrac{1-\sin2y}{\cos2y}$$

OR

$$(ii)=\dfrac{\cos^2y-\sin^2y}{(\cos y+\sin y)^2}=\dfrac{\cos2y}{1+\sin2y}$$

answered Nov 18 '17 at 18:50

lab bhattacharjee

274,582

So you mean choices are NOT correct??!! – Hamid Reza Ebrahimi Nov 18 '17 at 18:53
@HamidRezaEbrahimi, I have replaced $x=2y$ for the ease of writing. What is $(3)?$ – lab bhattacharjee Nov 18 '17 at 18:54

haqnatural · Accepted Answer · 2017-11-18T18:55:48.100

$$\frac { 1-\sin { x } +\cos { x } }{ 1+\sin { x } +\cos { x } } =\frac { \sin ^{ 2 }{ \left( \frac { x }{ 2 } \right) +\cos ^{ 2 }{ \left( \frac { x }{ 2 } \right) } -2\sin { \left( \frac { x }{ 2 } \right) } \cos { \left( \frac { x }{ 2 } \right) } +\cos ^{ 2 }{ \left( \frac { x }{ 2 } \right) } -\sin ^{ 2 }{ \left( \frac { x }{ 2 } \right) } } }{ \sin ^{ 2 }{ \left( \frac { x }{ 2 } \right) +\cos ^{ 2 }{ \left( \frac { x }{ 2 } \right) } +2\sin { \left( \frac { x }{ 2 } \right) } \cos { \left( \frac { x }{ 2 } \right) } +\cos ^{ 2 }{ \left( \frac { x }{ 2 } \right) } -\sin ^{ 2 }{ \left( \frac { x }{ 2 } \right) } } } =\\\ =\frac { 2\cos ^{ 2 }{ \left( \frac { x }{ 2 } \right) } -2\sin { \left( \frac { x }{ 2 } \right) } \cos { \left( \frac { x }{ 2 } \right) } }{ 2\cos ^{ 2 }{ \left( \frac { x }{ 2 } \right) } +2\sin { \left( \frac { x }{ 2 } \right) } \cos { \left( \frac { x }{ 2 } \right) } } =\\=\frac { \cos { \left( \frac { x }{ 2 } \right) } -\sin { \left( \frac { x }{ 2 } \right) } }{ \cos { \left( \frac { x }{ 2 } \right) } +\sin { \left( \frac { x }{ 2 } \right) } } =\frac { { \left( \cos { \left( \frac { x }{ 2 } \right) } -\sin { \left( \frac { x }{ 2 } \right) } \right) }^{ 2 } }{ \left( \cos { \left( \frac { x }{ 2 } \right) } +\sin { \left( \frac { x }{ 2 } \right) } \right) \left( \cos { \left( \frac { x }{ 2 } \right) } +\sin { \left( \frac { x }{ 2 } \right) } \right) } =\\ =\frac { 1-\sin { x } }{ \cos { x } } $$

@labbhattacharjee,yes, it was a typo, thank you for mentioning — haqnatural, Nov 18 '17 at 18:57

Guy Fsone · Answer 3 · 2017-11-19T17:09:21.783

First Method: Using the identities $$\cos a+ \cos b =2\cos\left(\frac{a+b}2\right)\cos\left(\frac{a-b}2\right)\tag{I}$$ and $$\cos a + \sin a =\sqrt2\cos\left(a-\frac{\pi}4\right)\tag{II}$$ we have, \begin{align}\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}&= \frac{1+\sqrt2\cos\left(x+\frac{\pi}4\right)}{1+\sqrt2\cos\left(x-\frac{\pi}4\right)} ~~~~\text{using (II)} \\&= \frac{\cos(\frac\pi4)+\cos\left(x+\frac{\pi}4\right)}{\cos(\frac\pi4)+\cos\left(x-\frac{\pi}4\right)} \\&=\frac{\cos\left(\frac{x}2+\frac{\pi}4\right)\cos\left(\frac{x}2\right)}{\cos\left(\frac{x}2-\frac{\pi}4\right)\cos\left(\frac{x}2\right)}~~~~\text{using (II)} \\&= \frac{2\cos\left(\frac{x}2+\frac{\pi}4\right)\cos\left(\frac{x}2-\frac{\pi}4\right)}{2\cos^2\left(\frac{x}2-\frac{\pi}4\right)} \\&=\frac{\cos\left(x\right)+\cos\left(\frac{\pi}2\right)}{1+\cos\left(\frac{\pi}2-x\right)}~~~~\text{using (I) and $\cos2x = 2\cos^2 x -1$ } \\&\color{red}{=\frac{\cos\left(x\right)}{1+\sin\left(x\right)}}~~~~\text{since}~~~~\cos( \frac{\pi}2-x) = \sin x \\&\color{red}{=\frac{\cos\left(x\right)}{1+\sin\left(x\right)} \frac{1-\sin\left(x\right)}{1-\sin\left(x\right)} = \frac{1-\sin\left(x\right)}{\cos\left(x\right)}}\end{align}

second Method: $$\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}\times\frac { \cos { x } }{ 1-\sin { x} } = \frac{\cos (x)+\cos^2(x)-\sin(x)\cos (x)}{(1+\sin(x))(1-\sin(x))\cos(x)(1-\sin(x))} \\= \frac{\cos (x)+\cos^2(x)-\sin(x)\cos (x)}{\cos(x)+\color{blue}{(1-\sin^2(x)}-\cos(x)\sin(x)} =\frac{\cos (x)+\cos^2(x)-\sin(x)\cos (x)}{\cos(x)+\color{blue}{\cos^2(x)}-\cos(x)\sin(x)} =\color{blue}{1} $$

Why did someone down vote this answer? And people it to be delete? — Guy Fsone, Nov 19 '17 at 17:59

score 0 · Answer 4 · answered Nov 18 '17 at 19:04

Multiply numerator and denominator by $1-\sin x+\cos x$: you get \begin{align} \frac{1-\sin x+\cos x}{1+\sin x+\cos x} &=\frac{(1-\sin x+\cos x)^2}{(1+\cos x)^2-\sin^2x} \\[6px] &=\frac{1+\sin^2x+\cos^2x-2\sin x+2\cos x-2\sin x\cos x}{1+2\cos x+\cos^2x-(1-\cos^2x)}\\[6px] &=\frac{2(1-\sin x+\cos x-\sin x\cos x)}{2\cos x(1+\cos x)}\\[6px] &=\frac{(1-\sin x)(1+\cos x)}{\cos x(1+\cos x)} \end{align}

Simplify $\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}$

4 Answers4