How do you solve:
(cot 4u-1/cot 4u+1)=(1-tan 4u/1+tan 4u)
2006-10-14 19:18:24 · 4 answers · asked by muffy_hlc 2 in Science & Mathematics ➔ Mathematics
L.H.S. = COT4U-1 / COT4U +1
= (1/ TAN4U)-1 / ( 1/TAN4U) +1 { Because 1/ tan4u = cot4u }
= 1 - TAN4U / 1 + TAN4U
= R.H.S.
YOUR PROBLEM IS SOLVED DEAR !
2006-10-14 19:24:48 · answer #1 · answered by Anand 1 · 0⤊ 0⤋
Don't let the 4u freak you out, note that by definition
cot z = 1/tan z , for a general z,
then the LHS(left hand side) becomes,
[(1/tan 4u) - 1] / [(1/tan 4u) +1]
= [(1-tan 4u) / tan 4u] / [(1+tan 4u) / tan 4u]
by making the denominator common for the top and bottom seperately.
Now note that whatever you do to both the top and bottom (i.e. same operation) of a fraction is the same as keeping the fraction as it is. Now use the same analogy in reverse.
You'll see easily that the tan 4u cancels out.
then you get,
(1-tan 4u) / (1+ tan 4u) on the RHS as required.
QED.
2006-10-15 04:22:01 · answer #2 · answered by yasiru89 6 · 0⤊ 0⤋
simply substitute 1/tan for cot or 1/cot for tan and cancel equals on each side of equation
2006-10-15 02:29:27 · answer #3 · answered by the shadow knows 3 · 0⤊ 0⤋
(cot4u-1)/(cot4u+1)=((1/tan4u)-1)/((1/tan4u)+1)=((1/tan4u)-(tan4u/tan4u))/((1/tan4u)+(tan4u/tan4u))=((1-tan4u)/tan4u)/((1+tan4u)/tan4u)=(1-tan4u)/(1+tan4u)
2006-10-15 02:40:31 · answer #4 · answered by Fateme A 1 · 0⤊ 0⤋