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1

Solved example of proving trigonometric identities

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

I. Choose what side of the identity are we going to work on

2

To prove an identity, we usually begin to work on the side of the equality that seems to be more complicated, or the side that is not expressed in terms of sine and cosine. In this particular case, we will choose to work on the left side $\frac{1}{\cos\left(x\right)}-\left(\frac{\cos\left(x\right)}{1+\sin\left(x\right)}\right)$ to reach the right side $\tan\left(x\right)$

II. Express in terms of sine and cosine

Applying the tangent identity: $\displaystyle\tan\left(\theta\right)=\frac{\sin\left(\theta\right)}{\cos\left(\theta\right)}$

$\frac{1}{\cos\left(x\right)}+\frac{-\cos\left(x\right)}{1+\sin\left(x\right)}=\frac{\sin\left(x\right)}{\cos\left(x\right)}$
3

Express both sides of the identity in terms of sine ($\sin(x)$) and cosine ($\cos(x)$)

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

III. Operate, group, simplify

4

Combine fractions with different denominator using the formula: $\displaystyle\frac{a}{b}+\frac{c}{d}=\frac{a\cdot d + b\cdot c}{b\cdot d}$

$\frac{1+\sin\left(x\right)-\cos\left(x\right)^2}{\cos\left(x\right)\left(1+\sin\left(x\right)\right)}=\frac{\sin\left(x\right)}{\cos\left(x\right)}$
5

Applying the trigonometric identity: $1-\cos\left(\theta\right)^2=\sin\left(\theta\right)^2$

$\frac{\sin\left(x\right)^2+\sin\left(x\right)}{\cos\left(x\right)\left(1+\sin\left(x\right)\right)}=\frac{\sin\left(x\right)}{\cos\left(x\right)}$
6

Factoring by $\sin\left(x\right)$

$\frac{\sin\left(x\right)\left(\sin\left(x\right)+1\right)}{\cos\left(x\right)\left(1+\sin\left(x\right)\right)}=\frac{\sin\left(x\right)}{\cos\left(x\right)}$
7

Simplify the fraction by $\sin\left(x\right)+1$

$\frac{\sin\left(x\right)}{\cos\left(x\right)}=\frac{\sin\left(x\right)}{\cos\left(x\right)}$

IV. Check if we arrived at the expression we wanted to prove

8

Since both sides of the equality are equal, we have proven the identity

true

Final Answer

true

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