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Proving Trigonometric Identities Calculator

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Example

Solved Problems

Difficult Problems

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

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

Any expression multiplied by $1$ is equal to itself

$\frac{1+\sin\left(x\right)-\cos\left(x\right)^2}{\cos\left(x\right)\left(1+\sin\left(x\right)\right)}=\frac{\sin\left(x\right)}{\cos\left(x\right)}$
4

Unifying fractions with different denominator

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

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

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

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

Any expression multiplied by $1$ is equal to itself

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

Multiplying polynomials $\cos\left(x\right)$ and $1+\sin\left(x\right)$

$\frac{\sin\left(x\right)^2+\sin\left(x\right)}{\cos\left(x\right)+\cos\left(x\right)\sin\left(x\right)}=\frac{\sin\left(x\right)}{\cos\left(x\right)}$
7

Apply fraction cross-multiplication

$\left(\sin\left(x\right)^2+\sin\left(x\right)\right)\cos\left(x\right)=\left(\cos\left(x\right)+\cos\left(x\right)\sin\left(x\right)\right)\sin\left(x\right)$

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

Any expression multiplied by $1$ is equal to itself

$\frac{\sin\left(x\right)^2+\sin\left(x\right)}{\cos\left(x\right)+\cos\left(x\right)\sin\left(x\right)}=\frac{\sin\left(x\right)}{\cos\left(x\right)}$
8

Multiplying polynomials $\cos\left(x\right)$ and $\sin\left(x\right)^2+\sin\left(x\right)$

$\cos\left(x\right)\sin\left(x\right)^2+\cos\left(x\right)\sin\left(x\right)=\left(\cos\left(x\right)+\cos\left(x\right)\sin\left(x\right)\right)\sin\left(x\right)$

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

Any expression multiplied by $1$ is equal to itself

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

$\cos\left(x\right)\sin\left(x\right)^2+\cos\left(x\right)\sin\left(x\right)=\sin\left(x\right)\cos\left(x\right)+\sin\left(x\right)\cos\left(x\right)\sin\left(x\right)$

When multiplying two powers that have the same base ($\sin\left(x\right)$), you can add the exponents

$\cos\left(x\right)\sin\left(x\right)^2+\cos\left(x\right)\sin\left(x\right)=\sin\left(x\right)\cos\left(x\right)+\sin\left(x\right)^2\cos\left(x\right)$
9

Multiplying polynomials $\sin\left(x\right)$ and $\cos\left(x\right)+\cos\left(x\right)\sin\left(x\right)$

$\cos\left(x\right)\sin\left(x\right)^2+\cos\left(x\right)\sin\left(x\right)=\sin\left(x\right)\cos\left(x\right)+\sin\left(x\right)^2\cos\left(x\right)$

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

10

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

true

Final Answer

true

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