tan 2 α + 2tan α . tan2 β = tan 2 β + 2tan β . tan2 α , if
Text Solution
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(b,c,d)
tan 2 α + 2 tan α . tan 2 β = tan 2 β + 2 tan β . tan 2 α
⇒ (tan 2 α – tan 2 β ) + 4 tan α tan β 
= 0
⇒ (tan 2 α – tan 2 β ) + 4 tan α tan β 
= 0
⇒ (tan 2 α – tan 2 β ) 
= 0 ⇒ (tan 2 α – tan 2 β ) (1 – tan 2 α . tan 2 β ) = 0
⇒ tan 2 α = tan 2 β or tan 2 α . tan 2 β = 1
L.H.S.= tan 2 α + 2 tan α 
. = tan 2 α + 
. (1 – tan 2 α ) = 1
R.H.S. = tan 2 β + 2 tan β . 
= tan 2 β + 
. (1 – tan 2 β ) = 1
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