## anonymous 3 years ago Verify (1/sinx+1) + (1/cscx+1) = 1

1. zepdrix

$\large \frac{1}{1+\sin x}+\frac{1}{\csc x+1}$ Let's do some manipulation. We're going to take advantage of two of the three basic square identities that you first learn in trig.$\large \color{royalblue}{\sin^2x+\cos^2x=1 \qquad \qquad \cot^2x+1=\csc^2x}$ In our problem, let's multiply the first term by the conjugate of the bottom.$\large \color{orangered}{\frac{1-\sin x}{1-\sin x}}\frac{1}{1+\sin x}+\frac{1}{\csc x+1}\color{orangered}{\frac{\csc x-1}{\csc x-1}}$ When we multiply conjugates, we end up with the Difference of squares on the bottoms.$\large \frac{1-\sin x}{1-\sin^2x}+\frac{\csc x-1}{\csc^2x-1}$

2. zepdrix

Now we can use our blue identities,$\large \color{royalblue}{\cos^2x=1-\sin^2x \qquad \qquad \cot^2x=\csc^2x-1}$ Using these identities gives us,$\large \frac{1-\sin x}{\color{royalblue}{1-\sin^2x}}+\frac{\csc x-1}{\color{royalblue}{\csc^2x-1}} \qquad \rightarrow \qquad \frac{1-\sin x}{\color{royalblue}{\cos^2x}}+\frac{\csc x-1}{\color{royalblue}{\cot^2x}}$

3. zepdrix

Recalling that, $\large \color{brown}{\cot^2x=\frac{\cos^2x}{\sin^2x}}$Gives us,$\large \frac{1-\sin x}{\cos^2x}+\frac{\csc x-1}{\color{brown}{\dfrac{\cos^2x}{\sin^2x}}}$Now notice that we're dividing by a fraction, which we can rewrite as multiplication if we flip the bottom fraction.$\large \frac{1-\sin x}{\cos^2x}+\frac{\color{brown}{\sin^2x}(\csc x-1)}{\color{brown}{\cos^2x}}$

4. zepdrix

Then distribute the sin^2x to each term in the brackets. There's only a little bit of simplification to do after that. Lemme know if you're still stuck on something.

5. anonymous

It took a while to read that since most of the formatting didn't work, but I think I've got it now. Thanks c:

6. zepdrix

Formatting didn't work? :c If you refresh the page it usually fixes it. That stinks :d

7. anonymous

I feel that the faster way of doing it without any identities is this. $\csc x=\frac{1}{\sin x}$ $\frac{ 1 }{ \sin x+1 }+\frac{ 1 }{ \csc x +1 }=\frac{ 1 }{ \sin x+1 }+\frac{ 1 }{ \dfrac{ 1 }{ \sin x } +1 }$ $\frac{ 1 }{ \sin x+1 }+\frac{ 1 }{ \csc x +1 }=\frac{ 1 }{ \sin x+1 }+\frac{ 1 }{ \dfrac{ 1+\sin x }{ \sin x } }$ $=\frac{ 1 }{ \sin x+1 }+\frac{ \sin x }{1+\sin x}$ $=\frac{1(1+\sin x+\sin x(\sin x +1)}{(\sin x+1)^2}$ $=\frac{1+\sin x+\sin^2 x +\sin x}{(\sin x +1)^2}$ $=\frac{\sin^2 x+2\sin x +1}{(\sin x +1)^2}$ Factorise the numerator $=\frac{(\sin x +1)^2}{(\sin x +1)^2}$ $=1$

8. anonymous

$1(1+\sin x+\sin x(\sin x+1)$ The numerator on one the lines was supposed to have a bracket somewhere. $1(1+\sin x)+\sin x(\sin x+1)$

9. anonymous

on one of*