do not use trigonometric tables.
Answers
LHS = tan 1 * tan 2 * tan 3 * .. tan89
= tan 1 * tan 2 * tan 3 * .. tan 45 *.. tan 87 * tan 88 * tan 89
= cot (90 - 1) * cot(90 - 2) * cot(90 - 3) *.. 1*.. tan 87 * tan 88 * tan 89
= cot 89 cot 88 cot 87 * .. * tan 87 tan 88 tan 89
Now we know cot x * tan x = 1/tan x * tan x = 1, so all cot and tan will cancel out and we get,
= 1
= RHS
(Pls mark as brainliest if you find it helpful :D)
Recall that
tan(90∘−x)=sin(90∘−x)cos(90∘−x)=cos(x)sin(x)=
cot(x)tan(90∘−x)=sin(90∘−x)cos(90∘−x)=c
os(x)sin(x)=cot(x)
Therefore
log(tan(1∘))+log(tan(89∘))=log(tan(1∘)tan(89∘))=
log(tan(1∘)cot(1∘))=log(1)=0log(tan(1∘))+log(tan(89∘))=
log(tan(1∘)tan(89∘))=log(tan(1∘)cot(1∘))=log(1)=0
and similarly
log(tan(2∘))+log(tan(88∘))=0log(tan(2∘))+log(tan(88∘))=0
log(tan(
2 tanθ=cot(90∘−θ)
∴tan1∘=cot89∘
cot89∘=tan89∘1
Similarly all the term will cancel out each other, except tan45∘, whose value is equal to 1.
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