F.4 Maths M2 question~

There could be some easier ways to do this, but this was how I did it...

LHS
= (1+csc(270+x)-tan(x-180)) / (1-csc(x-90)+tan(x+360))

Use the trigonometry identities e.g. sin(A+B) = ... equations,

tan(x-180) = tan(x+360) = tan x
csc(x-90) = csc(270+x) = -sec x

This gives LHS = (1 - sec x - tan x) / (1 + sec x + tan x)

= [ (1 - (sec x + tan x)) / (1 + (sec x + tan x)) ]

= [ (1 - (sec x + tan x)) / (1 + (sec x + tan x)) ] * [ (1 + (sec x + tan x) / (1 + (sec x + tan x) ]----multiply the top and bottom with the donorminator

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

Top: 1 -(sec x)^2 = -(tan x)^2; Bottom: combine the 1+(tan x)^2 = (sec x)^2

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

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

= - [tan x (tan x + sec x)] / [ (sec x +1) (sec x + tan x)]

= - tan x / (1 + sec x)

Almost there.

= - (sin x / cos x) / [ (cos x +1) / cos x]
= - sin x / (cos x + 1)

Multiply top and bottom by (cos x -1)
= - sin x (cos x - 1) / [ (cos x)^2 -1 ]

Notice: 1 - (cos x)^2 = (sin x)^2. Therefore (cos x)^2 - 1 = -(sin x)^2

= - sin x (cos x - 1) / -(sin x)^2
= - (cos x - 1) / -sin x

Divide top and bottom by cos x
= +(1 - sec x) / tan x
= RHS

上門補習:
名校中學畢業
學業成績曾試過在全班頭三名之內
教授 DSE 文憑數學 ( 基本課程，M1，M2）
每堂$ 500 (45-60 min)
第一，二堂(試堂) $0 . 免費 （只限兩堂，如試堂不適合可以不上第三堂 ）
詢問可電62806183


上門補習:
名校中學畢業
學業成績曾試過在全班頭三名之內
教授數學(中一至中六)
每堂$ 500 (45-60 min)
第一，二堂(試堂) $0 . 免費 （只限兩堂，如試堂不適合可以不上第三堂 ）
詢問可電62806183

It is quite difficult.
However, I find a easy way to solve the problem :)
First:
using the trigonometry identities to change
LHS to (1 - sec x - tan x) / (1 + sec x + tan x) .............(1)

Now, we need to prove (1 - sec x - tan x) / (1 + sec x + tan x)=(1-secx) / tanx

If we can prove (1 - sec x - tan x) =(1 + sec x + tan x)(1-secx) / tanx ,
the problem will be solved.

(1 + sec x + tan x)(1-secx) / tanx
=(1 + sec x + tan x - sec x - sec² x - tanx secx) / tan x
=((1 - sec² x) + tanx - tan x sec x )/tan x
= (-tan² x+ tanx - tan x sec x )/tan x
=1 - sec x -tanx
GOOD
Now, we can sub (1 - sec x - tan x) =(1 + sec x + tan x)(1-secx) / tanx into (1)

Q.E.D.