## Trigonometry (11th Edition) Clone

$$\frac{\sin s}{1+\cos s}+\frac{1+\cos s}{\sin s}=2\csc s$$ As 2 sides are proved equal below, the equation is an identity.
$$\frac{\sin s}{1+\cos s}+\frac{1+\cos s}{\sin s}=2\csc s$$ The left side, being more complex, would be dealt with first. $$X=\frac{\sin s}{1+\cos s}+\frac{1+\cos s}{\sin s}$$ $$X=\frac{\sin^2s+(1+\cos s)^2}{\sin s(1+\cos s)}$$ Recall that $(A+B)^2=A^2+2AB+B^2$, so we can expand $(1+\cos s)^2$ as follows. $$X=\frac{\sin^2s+1+2\cos s+\cos^2s}{\sin s(1+\cos s)}$$ $$X=\frac{(\sin^2s+\cos^2s)+1+2\cos s}{\sin s(1+\cos s)}$$ For $(\sin^2s+\cos^2s)$, it equals $1$, said by the Pythagorean Identities. $$X=\frac{1+1+2\cos s}{\sin s(1+\cos s)}$$ $$X=\frac{2+2\cos s}{\sin s(1+\cos s)}$$ $$X=\frac{2(1+\cos s)}{\sin s(1+\cos s)}$$ $$X=\frac{2}{\sin s}$$ Recall that $\frac{1}{\sin s}=\csc s$ (Reciprocal Identity) $$X=2\csc s$$ So, $$\frac{\sin s}{1+\cos s}+\frac{1+\cos s}{\sin s}=2\csc s$$ The equation is an identity, since 2 sides are proved equal.