1. 1+cos2A+cosA=cotA
sin2A+sinA
2. sin2A = tanA
1+cos2A
3. sin2A = cotA
1-cos2A
4. sin2A-cos2A=secA
sinA cosA
5. 1+sinA-cosA =tan A
1+sinA+cosA 2
THANK YOU!!
Hello, sanikui!
Here's #4 . . .
$\displaystyle 4)\;\;\frac{\sin2A}{\sin A} - \frac{\cos2A}{\cos A} \:=\:\sec A$
We have: .$\displaystyle \frac{\overbrace{2\sin A\cos A}^{\sin2A}}{\sin A} - \frac{\overbrace{2\cos^2\!A-1}^{\cos2A}}{\cos x} \;\;=\;\;\frac{2\sin A\cos A}{\sin A} - \frac{2\cos^2A}{\cos A} + \frac{1}{\cos A} $
. . $\displaystyle = \;\;2\cos A - 2\cos A + \frac{1}{\cos A}\;\;=\;\;\frac{1}{\cos A}\;\;=\;\;\sec A$
multiply denominator with both numer. and denom.
u will get
RHS=
$\displaystyle [1+1+2sin(A)-2cos(A) -2cos(A)sin(A)]$ / $\displaystyle [1+sin^2(A)+2sin(A)-cos^2(A)]$
numerator $\displaystyle =[2(1+sin(A))(1-cos(A)]$denominator $\displaystyle =[2sin(A)(1+sin(A)]$
cancel $\displaystyle 2(1+sin(A)$
=>RHS$\displaystyle = [1-cos(A)] / [sin(A)]$
Now u remember 3rd Question
u will get $\displaystyle 1/cot(A/2)= tan(A/2)$
dere may be shortcut for this!
sOrRy for I went offline ys'trday after 3rd