SOLUTION: I am a grade 10 students and i like trigonometry but i found this quiet tough.Plz can u solve it for me.. prove that (sec4a-1)/(sec8a-1)=tan4a/tan2a

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Question 61331: I am a grade 10 students and i like trigonometry but i found this quiet tough.Plz can u solve it for me..
prove that (sec4a-1)/(sec8a-1)=tan4a/tan2a
Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website!
I am a grade 10 students and i like trigonometry but i found this quiet tough.Plz can u solve it for me..
prove that (sec4a-1)/(sec8a-1)=tan4a/tan2a
FEW THINGS , I WANT TO KNOW..
1.IN 10 TH. GRADE ARE YOU TAUGHT MULTIPLE ANGLES.ASSUMING SO..
2.HAVE YOU TYPED THE PROBLEM CORRECTLY?IT DOES NOT APPEAR SO..GOR EXAMPLE TRY FOR A=15 DEGREES
LHS = [SEC(60)-1]/[SEC(120)-1]= 1/-3=-1/3
RHS = TAN(60)/TAN(30)= SQRT(3)/[1/SQRT(3)]=3...
SO THE PROBLEM IS NOT CORRECT....
3.ANY WAY LET US KEEP LHS IN TACT AND FIND IT OUT INTERMS OF TAN MULTIPLE ANGLES
LHS =[{1-C0S(4A)}/{COS(4A)}]/[{1-COS(8A)}/{COS(8A)}]
= 2SIN^2(2A)*COS(8A)/[2SIN^2(4A)COS(4A)]
=[{2SIN^2(2A)COS(8A)}]/[{2SIN(4A)COS(4A)}{SIN(4A)}]
=[2SIN^2(2A)COS(8A)]/[{SIN(8A)}{2SIN(2A)COS(2A)}]
= TAN(2A)/TAN(8A)....IS THE ANSWER
OK? GOT IT?.