# Proving trigonometric identities [duplicate]

I’ve had a bit of difficulty of this question: (1+sinA+cosA)/(1-sinA+cosA)=(1+sinA)/cosA

I tried to do: (SinA)^2+(CosA)^2+sinA+cosA/(SinA)^2+(CosA)^2-sinA+cosA=(1+sinA)/cosA But then I’m kind of lost. Any help will be appreciated! Additionally, I am not allowed to move one side to another (over the equal sign).

## marked as duplicate by lab bhattacharjee trigonometry StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 18 '18 at 16:47

• Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Nov 18 '18 at 16:37
• $$(1+s+c)c=(1+s)(1-s+c) \\c+cs+c^2=1-s+c+s-s^2+cs \\c^2=1-s^2$$ – Yves Daoust Nov 18 '18 at 16:48

\begin{align*} &\frac{1 + \sin A + \cos A}{1-\sin A + \cos A} = \frac{1 + \sin A}{\cos A} \\ &\iff \cos A + \sin A \cos A + \cos^2 A = 1 - \sin^2 A + \sin A \cos A + \cos A \\ &\iff \cos A + \sin A \cos A + \cos^2 A = \cos A + \sin A \cos A + \cos^2 A, \end{align*} where we used $$\sin^2 A + \cos^2 A = 1.$$

$$\frac{1+\sin A +\cos A}{1-\sin A + \cos A} = \frac{1+\sin A}{\cos A}\\ \cos A + \sin A \cos A + \cos^2 A = (1- \sin A + \cos A) (1+\sin A)\\ \cos A + \sin A \cos A + \cos^2 A = 1- \sin A + \cos A + \sin A - \sin^2 A + \sin A \cos A\\ \cos A + \sin A \cos A + \cos^2 A + \sin^2 A = 1- \sin A + \cos A + \sin A + \sin A \cos A\\ \cos A + \sin A \cos A + 1 = 1 + \cos A + \sin A \cos A\\$$

Asserting that$$\frac{1+\sin A+\cos A}{1-\sin A+\cos A}=\frac{1+\sin A}{\cos A}$$is equivalent to asserting that $$(1+\sin A+\cos A)\cos A=(1-\sin A+\cos A)(1+\sin A)$$. But$$(1+\sin A+\cos A)\cos A=\cos A+\sin(A)\cos(A)+\cos^2A$$and\begin{align}(1-\sin A+\cos A)(1+\sin A)&=(1-\sin A)(1+\sin A)+\cos A+\cos(A)\sin(A)\\&=1-\sin^2A+\cos A+\cos(A)\sin(A)\\&=\cos^2A+\cos A+\cos(A)\sin(A).\end{align}