Question

Prove that [sinx/1-cotx] + [cosx/1-tanx]=sinx+cosx

Answer 1

hmmm one sec

Answer 2

\(\bf \cfrac{sin(x)}{1-cot(x)}+\cfrac{cos(x)}{1-tan(x)}=sin(x)+cos(x)
\\ \quad \\ \quad \\
\cfrac{sin(x)}{1-cot(x)}+\cfrac{cos(x)}{1-tan(x)}\implies
\cfrac{sin(x)}{1-{\color{brown}{ \frac{cos(x)}{sin(x)}}}}+\cfrac{cos(x)}{1-{\color{brown}{ \frac{sin(x)}{cos(x)}}}}
\\ \quad \\
\cfrac{sin(x)}{\frac{sin(x)-cos(x)}{sin(x)}}+\cfrac{cos(x)}{\frac{cos(x)-sin(x)}{cos(x)}}
\\ \quad \\
\cfrac{sin(x)}{1}\cdot \cfrac{sin(x)}{sin(x)-cos(x)}+\cfrac{cos(x)}{1}\cdot \cfrac{cos(x)}{cos(x)-sin(x)}
\\ \quad \\
\cfrac{sin^2(x)}{sin(x)-cos(x)}+\cfrac{cos^2(x)}{cos(x)-sin(x)}
\\ \quad \\

\cfrac{sin^2(x)}{{\color{brown}{ -[cos(x)-sin(x)] }}}+\cfrac{cos^2(x)}{cos(x)-sin(x)}
\\ \quad \\
\cfrac{sin^2(x)-cos^2(x)}{-[cos(x)-sin(x)]}\implies
\cfrac{sin^2(x)-cos^2(x)}{sin(x)+cos(x)}\)

can you see where that goes?

Answer 3

Yes that is really helpful. Thank you!

Answer 4

so.. the numerator, is just a difference of squares :)
expand that, and the denominator goes kaput

Answer 5

yw