Question

Trig help (Problems already solved)?

Answer 1

The first identity uses the definition of the reciprocal functions $\sec x,\csc x,\cot x$ and the distributive property of multiplication.

$(\sec x+\sin x)\cot x=\left(\dfrac1{\cos x}+\sin x\right)\dfrac{\cos x}{\sin x}$
$=\dfrac{\cos x}{\cos x\sin x}+\dfrac{\cos x\sin x}{\sin x}$
$=\dfrac1{\sin x}+\cos x$
$=\csc x+\cos x$

The second uses the definition of $\tan x$ and the distributive property. Then a factor of $\dfrac1{\cos x}$ is pulled out, which allows you to use the identity $\sin^2x+\cos^2x=1$.

$\cos x+\tan x\sin x=\cos x+\dfrac{\sin x}{\cos x}\sin x$
$=\cos x+\dfrac{\sin^2x}{\cos x}$
$=\dfrac{\cos^2x}{\cos x}+\dfrac{\sin^2x}{\cos x}$
$=\dfrac1{\cos x}\left(\cos^2x+\sin^2x\right)$
$=\dfrac1{\cos x}\times1$
$=\dfrac1{\cos x}$
$=1$

The third uses the same ideas as the second: rewrite the reciprocal functions, then invoke the Pythagorean identity $\sin^2x+\cos^2x=1$, which is equivalent to $\sin^2x=1-\cos^2x$.

$\csc x-\cos x\cot x=\dfrac1{\sin x}-\cos x\dfrac{\cos x}{\sin x}$
$=\dfrac1{\sin x}-\dfrac{\cos^2x}{\sin x}$
$=\dfrac1{\sin x}\left(1-\cos^2x\right)$
$=\dfrac1{\sin x}\sin^2x$
$=\dfrac{\sin^2x}{\sin x}$
$=\sin x$

In the last one, you combine the fractions by enforcing common denominators. This lets you add the numerators together, and the denominator can be simplified. Once you do that, you rewrite the factors of cos and sin in the numerator and denominator to make up the cot and csc functions, and you're done.

$\dfrac{\cos x}{1+\cos x}+\dfrac{\cos x}{1-\cos x}=\dfrac{\cos x(1-\cos x)}{(1+\cos x)(1-\cos x)}+\dfrac{\cos x(1+\cos x)}{(1-\cos x)(1+\cos x)}$
$=\dfrac{\cos x(1-\cos x)+\cos x(1+\cos x)}{(1-\cos x)(1+\cos x)}$
$=\dfrac{\cos x(1-\cos x+1+\cos x)}{1-\cos^2x}$
$=\dfrac{2\cos x}{\sin^2x}$
$=2\dfrac{\cos x}{\sin x}\dfrac1{\sin x}$
$=2\cot x\csc x$