i'm having trouble getting this one started please.
Simplify the first trigonometric expression by writing the simplified form in terms of the second expression.
$$\frac{\sec x + \csc x}{1 + \tan x} \qquad \sin x$$
I have tried converting to
$$\frac{\dfrac{1}{\cos x} + \dfrac{1}{\sin x}}{1 + \dfrac{\sin x}{\cos x}}$$
Then
$$\frac{ 1 + \dfrac{\cos x}{\sin x}}{\cos x + \sin x}$$
But if I progress this further I cannot seem to yield a result and I'm not sure if I'm heading in the right direction witht this.
The answer is apparently, $\dfrac{1}{\sin x}$
Best Answer
We can rewrite numerator as,
$ \displaystyle \csc x + \sec x = \csc x \cdot \left(1 + \frac{\sec x}{\csc x}\right)$
$ \displaystyle~~~~ = \csc x \left(1 + \frac{\sin x}{\cos x}\right) = \csc x \left(1 + \tan x\right)$
So, $\displaystyle \frac {\csc x + \sec x}{1 + \tan x} = \frac{1}{\sin x}$