I've been doing this problem for 5 hours now

[hedstrong15]

csc(x) + sec(x) / sin(x) + cox(x) = cot(x) + tan(x)

I can't seem to verify it.
I reduced everything to cos and sin, I tried squaring, multiplying random identities ect. I don't know what to do.

 

[galactus]

\(\displaystyle \frac{\frac{1}{sin(x)}+\frac{1}{cos(x)}}{sin(x)+cos(x)}\)

\(\displaystyle =\frac{\frac{cos(x)+sin(x)}{sin(x)cos(x)}}{sin(x)+cos(x)}\)

\(\displaystyle =\frac{1}{sin(x)cos(x)}\)

Now, replace the numerator with \(\displaystyle cos^{2}(x)+sin^{2}(x)=1\)

\(\displaystyle \frac{cos^{2}(x)+sin^{2}(x)}{sin(x)cos(x)}\)

\(\displaystyle \frac{cos^{2}(x)}{sin(x)cos(x)}+\frac{sin^{2}(x)}{sin(x)cos(x)}\)

\(\displaystyle \frac{cos(x)}{sin(x)}+\frac{sin(x)}{cos(x)}=cot(x)+tan(x)\)

 

[hedstrong15]

Oh thank you so much. I never even though to multiply the top part of the equation only by the bases to make the denominator of the whole cancel. That is genius. Thank you so so much.

I also understand you can work both sides to make them = 1/sinxcosx, correct?