# How do you simplify (tan x + cos x +sin x tan x)/ (sec x + tan x)?

May 28, 2015

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

There are (many) others, but these are the most fundamental for trigonometry. In fact, every simple trigonometric function can be written in terms of sines and cosines. For example, the scary-looking "half coversed cosine" hacovercos(x) (which is not usually examinable) is simply $\frac{1 + \sin \left(x\right)}{2}$.

Using the above identities, suddenly your equation becomes:

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

From this point, it's algebraic manipulation and not much else more.

$= \frac{\sin \frac{x}{\cos} \left(x\right) + \cos \left(x\right) + \frac{{\sin}^{2} \left(x\right)}{\cos} \left(x\right)}{\frac{1 + \sin \left(x\right)}{\cos} \left(x\right)}$ (multiplication of the term in the numerator, and simplification of the denominator)
$= \frac{\sin \left(x\right) + {\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right)}{1 + \sin \left(x\right)}$ (multiplying top and bottom by $\cos \left(x\right)$)
=(sin(x)+1)/(1+sin(x) (because ${\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right) = 1$)
$= 1.$