This is an example of what's known as an identity, i.e., an equation that is true for all values of x. You don't "solve" it for x, but rather have to show that the two sides of the equation are always the same. We'll work with the left side of the equation and show that it's the same as the right side. The left side of the equation is the product of two binomials, in the form (a+b)(c+d). Using basic algebra this be expanded as ac+ad+bc+bd. When you apply this to the given formula, you get: 1 + 1/cos(x) - cos(x) - 1 which simplifies to 1/cos(x) - cos(x) Next you want a common denominator, so multiply cos(x) term by cos/cos (you can always multiply something by one) to yield: cos^2(x)/cos(x). So now the left side can be written as: (1 - cos^2(x))/cos(x) Now, there's an important trig identity that says that sin^2 + cos^2 =1, or in another form, 1 - cos^2(x) = sin^2(x). Right? So now the left side becomes: sin^2(x)/cos(x) This can be rearranged as: sin(x)*(sin(x)/cos(x)) But since sin/cos = tan, you can rewrite it as: sin(x)tan(x), which is exactly what you were looking for on the right side of the original equation. DONE!