Look at the attachment Simplified it too pls take it seriously

Tei asked Jan 15, 2023

188,040 views

4 votes

Look at the attachment
Simplified it too
pls take it seriously

Tei asked Jan 15, 2023

by Tei

3.0k points

Please log in or register to add a comment.

8 votes

$\\ \rm\rightarrowtail (sin(180+A)+2cos(180+A).cos(A-180))/(2cos^2(360+A)-cos(-A))$

$\\ \rm\rightarrowtail (-sinA+2(cos^2A-sin^2A))/(2cos^2A-cosA)$

$\\ \rm\rightarrowtail (-sinA+2cos^2A-2sin^2A)/(cosA(2cosA-1))$

$\\ \rm\rightarrowtail (-sinA+2-2sin^2A-2sin^2A)/(cosA(2cosA-1))$

$\\ \rm\rightarrowtail (-sinA-4sin^2A)/(cosA(2cosA-1))$

$\\ \rm\rightarrowtail (-sinA(4sinA-1))/(cosA(2cosA-1))$

$\\ \rm\rightarrowtail -tanA\left((4sinA-1)/(2cosA-1)\right)$

3.6k points

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

3.2m questions

4.1m answers