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sabika13
Group Title
How do I solve this using trig identities: tan(pi/4 + x) +tan(pi/4  x) =2sec2x
 one year ago
 one year ago
sabika13 Group Title
How do I solve this using trig identities: tan(pi/4 + x) +tan(pi/4  x) =2sec2x
 one year ago
 one year ago

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sabika13 Group TitleBest ResponseYou've already chosen the best response.0
i tried left side: 1+tanx+1tanx =2 :S
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
have you studied double angle identities?
 one year ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.0
well i have the formulas for it
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
ok good tell me formula for tan(a+b) and tan (ab) remember a and b are angles.
 one year ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.0
tan(a+/b)=(tana+/tanb)/(1+/tanatanb)
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
good.. now try to use this formula in your question.. it would be some thing like this for tan(pi/4 +x) = [tan(pi/4) +tanx]/(1tan(pi/4)tanx)
 one year ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.0
and then do i conjugate?
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
yes.. basically tan(pi/4) = 1 so just start solving as pi/4 will be eliminated.. the rest would be LCM and simple algebra.
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
i think it should work.. as where 2 angles are involved we use double angle identities to make stuff easy for us.. so this should work out.
 one year ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.0
i ended up with tanx
 one year ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.0
actually it was 1+tan^2x/1+tanx
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
seriously.. hmm i will solve it and will see.. i have to go right now but will be back in a while just post the working u did.. if i can do i will post back..
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
hmm i dont think so you did right.. as far as i know there should be 1tan^2x in the denominator.
 one year ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.0
i got (1+tanx)(1tanx)in deno
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
yes.. multiplying it would give (1tan^2x)
 one year ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.0
okaylet me show what i did: tanpi/4 + tanx)^2/(1tanpi/4tanx)(1+tanpi/4tanx) + tanpi/4tanx/1+tanpi/4+tanx
 one year ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.0
(tanpi/4+tanx)^2 + (tanpi/4tanx) / (1tanpi/4tanx)(1+tanpi/4tanx)
 one year ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.0
and then tanpi/4 =1
 one year ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.0
2+tan^2xtanx / 1tan^2x
 one year ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.0
ru there?
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
yes i am.. just solving for you..
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
ok well fault i found in your answer is in frist step. tanpi/4 + tanx)^2 how you got the square here?
 one year ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.0
i multiplied (tan pi/4 +tanx) to both top and bottom
 one year ago

sabika13 Group TitleBest ResponseYou've already chosen the best response.0
it will become a common deno
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
ok i got easy way.. try wolfram.. http://www.wolframalpha.com/input/?i=tan%28pi%2F4+%2B+x%29+%2Btan%28pi%2F4++x%29+%3D2sec%282x%29&dataset=&equal=Submit
 one year ago

nubeer Group TitleBest ResponseYou've already chosen the best response.0
tell me if this helps you.
 one year ago
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