plz....please help me! URGENT!!

1)
LHS
=1/(1-tanx) - 1/(1+tanx)
= [(1+tanx) - (1-tanx)] / (1-tanx)(1+tanx) [通分母]
= 2 tanx / (1-tan² x) [因為a²-b²=(a-b)(a+b)]
= tan (2x) [Double Angle Formula]
=RHS

2)
LHS
=cos(4x)
=cos(2*2x)
=2[cos(2x)]² - 1 [Double Angle Formula]
=2[2cos² x -1]² -1
=2[4(cos x)^4 - 4cos²x + 1] - 1
=8(cos x)^4 - 8 cos²x +1
=RHS

3)
sin 5 A / sin A - cos 5 A / cos A
= [sin(5A)cos(A) - cos(5A)sin(A)]/(sin(A)cos(A)) [通分母]
= sin(5A-A)/(sin(A)cos(A)) [Compound Angle Formula]
= sin(4A)/(sin(A)cos(A))
= sin(2*2A)/(sin(A)cos(A))
= 2sin(2A)cos(2A)/(sin(A)cos(A))
= 4sin(A)cos(A)cos(2A)/(sin(A)cos(A))
= 4cos(2A) [上下約簡]
= 4(1-2sin²A) [Double Angle Formula]
= 4 - 8sin²A
=RHS

希望幫倒你！ ^^

1)
1/1-tanx - 1/1+tanx = tan 2x
(1+tanx)/[(1-tanx)(1+tanx)] - (1-tanx)/[(1-tanx)(1+tanx)] = tan 2x
[(1+tanx)-(1-tanx)]/(1-tan^2x) = tan 2x
2tanx/(1-tan^2x) = tan 2x
tan 2x= tan 2x

2)
cos 4 x = 8 cos ^4 x - 8 cos ^2 x +1
cos(2x+2x)=8 cos ^4 x - 8 cos ^2 x +1
cos(2x) * cos(2x) - sin(2x) * sin(2x) =8 cos ^4 x - 8 cos ^2 x +1
cos^2 (2x) - sin^2 (2x) = 8 cos ^4 x - 8 cos ^2 x +1
cos^2 (2x) - [ 1-cos^2 (2x)] =8 cos ^4 x - 8 cos ^2 x +1
2cos^2(2x) - 1 =8 cos ^4 x - 8 cos ^2 x +1
2 cos (2x) * cos(2x) - 1 =8 cos ^4 x - 8 cos ^2 x +1
2 (cos x*cos x - sin x*sin x)*(cos x * cos x - sin x * sin x)- 1 = 8 cos ^4 x - 8 cos ^2 x +1
2 (cos^2 x - sin^2 x)*(cos^2 x - sin^2 x)-1 =8 cos ^4 x - 8 cos ^2 x +1
2 [cos^2 x - (1 - cos^2 x)]*[cos^2 x - (1 - cos^2 x)] - 1 =8 cos ^4 x - 8 cos ^2 x +1
2(2cos^2x-1)*(2cos^2x-1) - 1 =8 cos ^4 x - 8 cos ^2 x +1
8 cos ^4 x - 8 cos ^2 x +1=8 cos ^4 x - 8 cos ^2 x +1


3)
sin 5 A / sin A - cos 5 A / cos A = 4 - 8 sin ^2 A
the left hand side
=sin( 4 A +A)/ sin A - cos (4 A+A) / cos A
=[sin(4A)cosA+cos(4A)sinA]/sinA - (cos4AcosA-sin4AsinA)/cosA
=[sin(2A+2A)cosA+cos(2A+2A)sinA]/sinA - [cos(2A+2A)cosA-sin(2A+2A)sinA]/cosA
=[2sin(2A)cos(2A)cosA + (1-2sin^2(2A))sinA]/sinA - [(1-2sin^2(2A))cosA - 2sin(2A)cos(2A)sinA]/cosA
and so on expand then simplifiy u will get the answer

X+6
cos 5x
6a