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Question

P-1:-tan1.tan2.tan3......tan87.tan88.tan89=1


Solution

tan 1 . tan 2 . tan 3 ... tan 87 . tan 88 . tan 89 = LHS
tan 1 . tan 2 . tan 3 ... tan (90 - 3 ) . tan ( 90 - 2 ) . tan ( 90 - 1)
tan 1 . tan 2 . tan 3 ... cot 3 . cot 2 . cot 1 
since tan(90-x)=cotx
tan 1 . cot 1 . tan 2 . cot 2 . tan 3 . cot 3 ... tan 89 . cot 89
1 x 1 x 1 x 1 x ... x 1
As 1ⁿ = 1 = RHS
Where n = any positive integer

OR
Alternatively,
tan1.tan2......tan45.....tan88.tan89

tan1.tan2....tan45...tan(90-2).tan(90-1)

tan1.tan2....tan45....cot2.cot1

We know that :- tanA.cotA=1

so ,1^44.tan45

= 1*1

=1

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