1. ## How to prove?trigo

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!!

2. ## 1st question!

Hello !
cos(2A)=cos^2(A)-sin^2(A)
sin(2A)=2sin(A)cos(A)
put in the above to get
[2cos^2 (A)+cos(A)] / [2sin(A)cos(A)+sin(A)]
Take cos(A) from numerator
& sin(A) from denominatorcommon you get cot(A)

3. ## 2nd & 3rd

put
1+cos(2A)=2cos^2(A) for 2nd denom.

1-cos(2A)=2sin^2(A) for 3rd denom.

sin(2A)=2sin(A)cos(A) for numerator

and cancel common terms.

4. k...i got it....thx

5. Hello, sanikui!

Here's #4 . . .

$\displaystyle 4)\;\;\frac{\sin2A}{\sin A} - \frac{\cos2A}{\cos A} \:=\:\sec A$

We have: .$\displaystyle \frac{\overbrace{2\sin A\cos A}^{\sin2A}}{\sin A} - \frac{\overbrace{2\cos^2\!A-1}^{\cos2A}}{\cos x} \;\;=\;\;\frac{2\sin A\cos A}{\sin A} - \frac{2\cos^2A}{\cos A} + \frac{1}{\cos A}$

. . $\displaystyle = \;\;2\cos A - 2\cos A + \frac{1}{\cos A}\;\;=\;\;\frac{1}{\cos A}\;\;=\;\;\sec A$

6. ## 5th one

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

7. Thx ADARSH and Soroban,i try my best~!

8. Question 5
i think no nid multiply denominator with both numer and denom.
Just use half-angles formulas...then can do it.