tan 2x/1+sec 2x = tanx
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HEYA,
HERE IS YOUR ANSWER
(tan2x)/(1+sec2x))= tanx
LHS =(tan2x)/(1+sec2x))
{sin2x/cos2x}/{1+1/cos2x} Since sec2x=1/cos2x
={sin2x/cos2x}/{(cos2x+1)/cos2x} Taking LCM
={(sin2x)/(cos2x+1).......... Canceling out cos2x
=(2sinxcosx)/(2cos²x−1+1).......Using cos2x=2cos²x−1
=(sinxcosx)/(cos²x)..........Canceling out cosx
=(sinx)/(cosx)=tanx=RHS
HERE IS YOUR ANSWER
(tan2x)/(1+sec2x))= tanx
LHS =(tan2x)/(1+sec2x))
{sin2x/cos2x}/{1+1/cos2x} Since sec2x=1/cos2x
={sin2x/cos2x}/{(cos2x+1)/cos2x} Taking LCM
={(sin2x)/(cos2x+1).......... Canceling out cos2x
=(2sinxcosx)/(2cos²x−1+1).......Using cos2x=2cos²x−1
=(sinxcosx)/(cos²x)..........Canceling out cosx
=(sinx)/(cosx)=tanx=RHS
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Answered by
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Answer:
Hence proved
Step-by-step explanation:
(tan2x)/(1+sec2x))= tanx
LHS =(tan2x)/(1+sec2x))
{sin2x/cos2x}/{1+1/cos2x} Since sec2x=1/cos2x
={sin2x/cos2x}/{(cos2x+1)/cos2x} Taking LCM
={(sin2x)/(cos2x+1).......... Canceling out cos2x
=(2sinxcosx)/(2cos²x−1+1).......Using cos2x=2cos²x−1
=(sinxcosx)/(cos²x)..........Canceling out cosx
=(sinx)/(cosx)=tanx=RHS
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