# Prove that $(\csc A+\cot A)\,\text{covers}\,A-(\sec A+\tan A)\,\text{vers}\,A=(\csc A-\sec A)(2-\text{vers}\,A\,\text{covers}\,A)$

Prove that the following equation is true. $$\newcommand{covers}{\operatorname{covers}}\newcommand{vers}{\operatorname{vers}} (\csc A+\cot A)\covers A-(\sec A+\tan A)\vers A=(\csc A-\sec A)(2-\vers A\covers A)$$ where $$\vers x=1-\cos x\\\covers x=1-\sin x.$$

This is in Plane Trigonometry by SL Loney. Please explain how I can solve it according to class 11th student. My working is below:

\begin{align}\text{LHS}&=(\csc A+\cot A)\covers A-(\sec A+\tan A)\vers A\\&=(\csc A+\cot A)(1-\sin A)-(\sec A+\tan A)(1-\cos A)\\&=\csc A+\cot A-1-\cos A-(\sec A+\tan A-1-\sin A)\\&=\frac1{\sin A}+\frac{\cos A}{\sin A}-\cos A-\frac 1{\cos A}-\frac{\sin A}{\cos A}+\sin A\\&=\frac{1+\cos A+\sin^2A+1-1}{\sin A}-\frac{1+\sin A+\cos^2A+1-1}{\cos A}\\&=\frac{2+\cos A-\cos^2A}{\sin A}-\frac{2+\sin A-\sin^2A}{\cos A}\end{align}

but I don't know how to continue.

• Use MathJAX instead of pictures. Commented May 26, 2019 at 18:36
• Here's a nice mathjax tutorial: math.meta.stackexchange.com/questions/5020/… Commented May 26, 2019 at 18:40
• How is it possible for me, during 50+ years of mathematical activity, not to have been in contact with these wonderful "vers" and "covers" functions :) Commented May 26, 2019 at 20:05

Rewriting in more common trigonometric functions, the statement is equivalent to $$\small\left(\frac1{\sin A}+\frac{\cos A}{\sin A}\right)(1-\sin A)-\left(\frac1{\cos A}+\frac{\sin A}{\cos A}\right)(1-\cos A)=\left(\frac1{\sin A}-\frac1{\cos A}\right)\left(2-(1-\sin A)(1-\cos A)\right)$$ which can be simplified to \begin{align}\small\frac{(1+\cos A)(1-\sin A)}{\sin A}-\frac{(1-\cos A)(1+\sin A)}{\cos A}&\small=(1+\cos A+\sin A-\sin A\cos A)\left(\frac1{\sin A}-\frac1{\cos A}\right)\\&\small=\left(\frac{1+\cos A}{\sin A}+1-\cos A\right)-\left(\frac{1+\sin A}{\cos A}+1-\sin A\right)\\&\small=\sin A-\cos A+\frac{1+\cos A}{\sin A}-\frac{1+\sin A}{\cos A}\end{align} and moving the fractions across to the left leads to \begin{align}\text{LHS}&=\frac{1+\cos A}{\sin A}(1-\sin A-1)-\frac{1+\sin A}{\cos A}(1-\cos A-1)\\&=-(1+\cos A)-[-(1+\sin A)]\\&=\sin A-\cos A=\text{RHS}\end{align} so the statement is true.