pls also try this... (tanx/1-cotx) + (cotx/1-tanx) = 1 + tanx - cotx
2006-12-04 18:31:18 · 4 answers · asked by justanasker 1 in Science & Mathematics ➔ Mathematics
Not sure bout the first one but I got the second one:( I think yur RHS shud be "1+tanx+cotx"
tanx/[1-(1/tanx)] +cotx/(1-tanx)
= (tanx)(tanx)/tanx-1 +cotx/(1-tanx)
=tanx(tanx)/tanx-1 - cotx/tanx-1
={[tanx(tanx)] - cotx}/ tanx- 1
its really hard to write the rest down so I'll try and explain it. So u got "[tanx(tanx)] - cotx"
the whole divided by "tanx -1"
u write "cotx" as '1/tanx' . then u'll get
=tanx(tanx)(tanx) - 1/ (tanx-1)tanx
the numerator is in the form of the identity (a3-b3) = (a-b)(a2+ab+b2)
u expand it and get
(tanx-1)(tanx2 +tanx +1) / (tanx-1)tanx
= tan2x +tanx+1/ tanx
divide each term of the numerator by "tanx " and you get
"1 +tanx + cotx"
Im sorry its hard to understand but hope u got sumthin :)
2006-12-04 19:01:11 · answer #1 · answered by Lynne 4 · 0⤊ 0⤋
(secx-tanx)^2+1
= Secx^2 - 2.secx.tanx + Tanx^2 + 1
[ Tanx^2 + 1 = Secx^2 ]
= 2Secx^2 - 2.Secx.tanx
=2Secx [ Secx - Tanx]
Therefore the EQuation u gave BEcomes
2Secx [ Secx - tanx ] / [Secx - Tanx ]
= 2Secx
As LHS = RHS Hence Proved
Signing off.... Vivek
2006-12-04 18:53:07 · answer #2 · answered by vedic_vivek 1 · 0⤊ 0⤋
1)LHS=(secx-tanx)² + 1 / secx-tanx
We know that sec²x - tan²x=1
Substitute this in the numerator
=>[(secx - tanx)² + (sec²x - tan²x)]/secx-tanx
=>[(secx - tanx)² + (secx - tanx)(secx + tanx)]/secx-tanx
taking (secx-tanx ) comman in the numerator
=>(secx-tanx) [secx-tanx + secx + tanx]/secx-tanx
=>2secx=RHS
Hence proved.
2)LHS=(tanx/1-cotx) + (cotx/1-tanx)
put cotx=1/tanx
=>tanx/(1-1/tanx) + (1/tanx)/(1-tanx)
=>tan²x/(tanx - 1) + 1/[(1-tanx)tanx]
taking LCM
=>[-tan²x(tanx) + 1]/(1-tanx)tanx
=>[1-(tanx)^3]/(1-tanx)tanx
now a^3 - b^3 =(a-b)(a^2 + b^2 + ab)
=>(1-tanx)(1+ tan²x + tanx)/(1-tanx)tanx
=>(1+tan²x + tanx)/tanx
=>cotx + tanx +1
=>The RHS term given by you is wrong, the correct form is as calculated!
LHS, RHS denote left hand side, right hand side
2006-12-05 01:12:45 · answer #3 · answered by sushant 3 · 0⤊ 0⤋
sec^2x+tan^2x-2secxtanx+1/(secx-tanx)
As 1+tan^2x=sec^2x, replacing it in the above equation
sec^2x+sec^2x-2secxtanx/(secx-tanx)
2sec^2x-2secxtanx/(secx-tanx)
2secx(secx-tanx)/(secx-tanx)
2secx
2006-12-04 19:24:03 · answer #4 · answered by raghunandan r 1 · 0⤊ 0⤋