$\frac{1-\tan\left(x\right)}{1+\tan\left(x\right)}$

Step-by-step Solution

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Final answer to the problem

$\frac{\left(1-\tan\left(x\right)\right)\cos\left(x\right)}{\sin\left(x\right)+\cos\left(x\right)}$
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Step-by-step Solution

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Apply the trigonometric identity: $\tan\left(\theta \right)$$=\frac{\sin\left(\theta \right)}{\cos\left(\theta \right)}$

$\frac{1+\frac{-\sin\left(x\right)}{\cos\left(x\right)}}{1+\frac{\sin\left(x\right)}{\cos\left(x\right)}}$
Why is tan(x) = sin(x)/cos(x) ?

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

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Learn how to solve problems step by step online. (1-tan(x))/(1+tan(x)). Apply the trigonometric identity: \tan\left(\theta \right)=\frac{\sin\left(\theta \right)}{\cos\left(\theta \right)}. Apply the formula: a+\frac{b}{c}=\frac{b+ac}{c}, where a=1, b=\sin\left(x\right), c=\cos\left(x\right), a+b/c=1+\frac{\sin\left(x\right)}{\cos\left(x\right)} and b/c=\frac{\sin\left(x\right)}{\cos\left(x\right)}. Apply the formula: \frac{a}{\frac{b}{c}}=\frac{ac}{b}, where a=1+\frac{-\sin\left(x\right)}{\cos\left(x\right)}, b=\sin\left(x\right)+\cos\left(x\right), c=\cos\left(x\right), a/b/c=\frac{1+\frac{-\sin\left(x\right)}{\cos\left(x\right)}}{\frac{\sin\left(x\right)+\cos\left(x\right)}{\cos\left(x\right)}} and b/c=\frac{\sin\left(x\right)+\cos\left(x\right)}{\cos\left(x\right)}. Apply the trigonometric identity: \frac{\sin\left(\theta \right)}{\cos\left(\theta \right)}=\tan\left(\theta \right).

Final answer to the problem

$\frac{\left(1-\tan\left(x\right)\right)\cos\left(x\right)}{\sin\left(x\right)+\cos\left(x\right)}$

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Plotting: $\frac{\left(1-\tan\left(x\right)\right)\cos\left(x\right)}{\sin\left(x\right)+\cos\left(x\right)}$

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4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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