2
$\begingroup$

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).

$\endgroup$

marked as duplicate by lab bhattacharjee trigonometry Nov 18 '18 at 16:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

\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 $\sin^2 A + \cos^2 A = 1.$

$\endgroup$
1
$\begingroup$

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

Read from bottom to top.

$\endgroup$
1
$\begingroup$

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}

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.