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Question
Prove that (1 + tan 1°)(1 + tan 2°)(1 + tan 3°) ..... (1 + tan 44°) is a multiple of 4
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Solution
1 + tan 44° = 1 + tan(45° – 1°)
= ` 1 + (tan 45^circ - tan 1^circ)/(1 - tan 45^circ tan 1^circ)`
= `1 + (1 - tan 1^circ)/(1 + tan 1^circ)`
= `(1 + tan 1^circ + 1 - tan 1^circ)/(1 - 1 tan 1^circ)`
= `2/(1 - 1 tan 1^circ)`
(1 + tan 1°)(1 + tan 44°) = 2
Similarly (1 + tan 2°)(1 + tan 43°) = 2
(1 + tan 3°)(1 + tan 42°) = 2
(1 + tan 22°)(1 + tan 23°) = 2
= (1 + tan 1°)(1 + tan 2°) … (1 + tan 44°)
= 2 × 2 × … 22 times
It is a multiple of 4.
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