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Exercise

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

Step-by-step Solution

1

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

$\begin{array}{l}L.C.M.=\left(1+\sin\left(x\right)\right)\left(1-\sin\left(x\right)\right) \\ L.C.M.=1-\sin\left(x\right)^2 \\ L.C.M.=\cos\left(x\right)^2\end{array}$
2

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

$\frac{1-\sin\left(x\right)}{\cos\left(x\right)^2}+\frac{1+\sin\left(x\right)}{\cos\left(x\right)^2}$
3

Combine and simplify all terms in the same fraction with common denominator $\cos\left(x\right)^2$

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

Final answer to the exercise

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

Try other ways to solve this exercise

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Similar Questions Solved

$\frac{4}{x+2}=\frac{6}{x-2}$

$\frac{3}{x+1}-\frac{1}{2}=\frac{1}{3x+3}$

$\frac{2x+4}{5x}=\frac{2}{x}$

$\frac{5}{x-7}=\frac{2}{x-2}$

$\frac{2x+3}{3x}=\frac{x}{2x-3}$

$\frac{x+1}{x-2}=\frac{x-3}{x}$

$\frac{x}{2}=\frac{8}{x}$

Main Topic: Equations

In mathematics, an equation is a statement of an equality containing one or more variables. Solving the equation consists of determining which values of the variables make the equality true. In this situation, variables are also known as unknowns and the values which satisfy the equality are known as solutions.

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