Find the simplified form of $\frac {1}{\cos x + \sin x}$ Find the simplified form of $\dfrac {1}{\cos x + \sin x}$.
a). $\dfrac {\sin (\dfrac {\pi}{4} +x)}{\sqrt {2}}$
b). $\dfrac {\csc (\dfrac {\pi}{4} + x)}{\sqrt {2}}$
c). $\dfrac {\sin (\dfrac {\pi}{4} + x)}{2}$
d). $\dfrac {\csc (\dfrac {\pi}{4} + x)}{2}$
My Attempt:
$$=\dfrac {1}{\cos x + \sin x}$$
$$=\dfrac {1}{\cos x + \sin x} \times \dfrac {\cos x - \sin x}{\cos x - \sin x}$$
$$=\dfrac {\cos x - \sin x}{\cos^2 x - \sin^2 x}$$
$$=\dfrac {\cos x - \sin x}{\cos 2x}$$


This is the graph of the function $\sin x+\cos x$.
Notice that itself looks like a wave.
So, we should be able to deduce that(since only the amplitude and phase seem to have changed),
$$\sin x+\cos x=A\sin( x+\phi)$$
Expanding using the identity for $\sin(A+B)$,
$$\sin x+\cos x=A\sin( x)\cos\phi+A\cos(x)\sin\phi$$
Comparing the coefficients of $\sin $ and $\cos,
$$1=A\cos\phi$$
$$1=A\sin\phi$$
Assuming non-zero $\cos$,dividing the second equation by the first gives:
$$1=\tan\phi\implies \phi=\frac{\pi}{4}$$
We only need one solution so we won"t worry about other solutions.
Now squaring the equations and adding,
$$2=A^2\implies A=\sqrt{2}$$
