If tan 20° = k, then tan 250° + tan 340°÷ tan 200° - tan 110° =
1) 1+k÷1-k
2) 1-k ÷1+k
3) 1 + k sq÷ 1+k sq
4) 1-k sq÷ 1+k sq
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Answered by
68
given
tan 20 = k----(1)
tan 250 = tan (270-20) = cot 20 = 1/tan 20 = 1/k
tan 340 = tan(360-20) = - tan 20 = -k
tan 200 = tan (180+20) = tan 20 =k
tan 110 = tan (90+20 ) = -cot 20 = -1/tan 20 = -1/k
(tan 250 +tan 340) /( tan200 -tan 110)
= (1/k -k)/(k + 1/k)
= (1 -k²)/ (k²+1)
4) is correct
tan 20 = k----(1)
tan 250 = tan (270-20) = cot 20 = 1/tan 20 = 1/k
tan 340 = tan(360-20) = - tan 20 = -k
tan 200 = tan (180+20) = tan 20 =k
tan 110 = tan (90+20 ) = -cot 20 = -1/tan 20 = -1/k
(tan 250 +tan 340) /( tan200 -tan 110)
= (1/k -k)/(k + 1/k)
= (1 -k²)/ (k²+1)
4) is correct
bhanusri7:
sorry your answer is correct I didn't mention sq for it.
thank you so much nice explanation .
u"r welcome
Thank u selecting as brainliest
:) it's OK .no problem .
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