How do you simplify #tanx + cosx/(1 + sinx)#?

Question

Lily Anderson · Accepted Answer

#(sinx(sinx + 1))/((cosx)(sinx + 1)) + ((cosx)(cosx))/((1 + sinx)(cosx))# #(sin^2x + sinx + cos^2x)/((cosx)(sinx + 1))# Applying the Pythagorean identity #sin^2x + cos^2x = 1#: #(1 + sinx)/((cosx)(sinx + 1))# #1/cosx# Applying the reciprocal identity #1/cosx = secx#: #secx# Hopefully this helps!

Sadie Alexander · Answer

undefined To simplify $ \frac{\tan(x) + \cos(x)}{1 + \sin(x)} $, we can use trigonometric identities to rewrite the expression.

1. First, recall that $ \tan(x) = \frac{\sin(x)}{\cos(x)} $.
2. Next, we know that $ \cos(x) = \frac{1}{\sec(x)} $ and $ \sin(x) = \frac{1}{\csc(x)} $.

Substitute these into the expression:

$$ \frac{\frac{\sin(x)}{\cos(x)} + \cos(x)}{1 + \frac{1}{\csc(x)}} $$

Now, simplify the expression:

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

So, $ \frac{\tan(x) + \cos(x)}{1 + \sin(x)} $ simplifies to $ \frac{1}{1 + \sin(x)} $.