show that tan(45+A) tan(45-A)=1
Answers
Answered by
76
tan(x+y ) = (tanx + tan y)/(1- tanx tany)
tan 45 = 1
lhs = tan(45+A)tan(45-A)
= (tan 45 +tan A)/(1- tan45tanA) * (tan45- tanA)/(1+tan45tanA)
= (1+tanA)/(1- tanA) * (1- tanA)/(1+tanA)
after cancellation
=1
RHS
tan 45 = 1
lhs = tan(45+A)tan(45-A)
= (tan 45 +tan A)/(1- tan45tanA) * (tan45- tanA)/(1+tan45tanA)
= (1+tanA)/(1- tanA) * (1- tanA)/(1+tanA)
after cancellation
=1
RHS
Answered by
32
LHS
= tan (45+A) tan (45-A)
= tan (45+A) cot [90-(45-A]
-----------------> {since tan A = cot (90-A)
= tan (45+A) cot (90-45+A)
= tan (45+A) × cot (45+A)
= 1
------------------> (since tanA.cotA = 1)
= RHS
-------> Hence proved
_________________________
= tan (45+A) tan (45-A)
= tan (45+A) cot [90-(45-A]
-----------------> {since tan A = cot (90-A)
= tan (45+A) cot (90-45+A)
= tan (45+A) × cot (45+A)
= 1
------------------> (since tanA.cotA = 1)
= RHS
-------> Hence proved
_________________________
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