how is (seca)(secb)(secc)=(tana+i)(tanb+i)(tanc+i)i=(-1)^1/2
how is (seca)(secb)(secc)=(tana+i)(tanb+i)(tanc+i)
i=(-1)^1/2
Himanshu Singhal , 10 Years ago
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Trigonometry> how is (seca)(secb)(secc)=(tana+i)(tanb+i...
1 Answers
noogler
Last Activity: 10 Years ago
it is true only if a+b+c=3pi/2
because , consider R.H.S
put Tana=sina/cosa, Tanb=sinb/cos, tanc=sinc/cosc
=[sina+icosa][sinb+icosb]sinc+icosc]/cosa.cosb.cosc
=cis(3pi/2-(a+b+c))[seca.secb.secc]
if a+b+c=3pi/2 then =cis(0)[seca.secb.secc]=seca.secb.secc
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