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1

Solved example of proving trigonometric identities

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

The least common multiple (LCM) of a sum of algebraic fractions consists of the product of the common factors with the greatest exponent, and the uncommon factors

$L.C.M.=\cos\left(x\right)\left(1+\sin\left(x\right)\right)$
3

Obtained the least common multiple, we place the LCM as the denominator of each fraction and in the numerator of each fraction we add the factors that we need to complete

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

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

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

Combine and simplify all terms in the same fraction with common denominator $\cos\left(x\right)\left(1+\sin\left(x\right)\right)$

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

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

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

Factor the polynomial $\sin\left(x\right)^2+\sin\left(x\right)$ by it's GCF: $\sin\left(x\right)$

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

Simplify the fraction $\frac{\sin\left(x\right)\left(\sin\left(x\right)+1\right)}{\cos\left(x\right)\left(1+\sin\left(x\right)\right)}$ by $\sin\left(x\right)+1$

$\frac{\sin\left(x\right)}{\cos\left(x\right)}=\tan\left(x\right)$
8

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

$\tan\left(x\right)=\tan\left(x\right)$
9

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

true

Final Answer

true

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