tan10°-tan50° tan70°
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Answered by
1
From the Log Table, we get
tan 10° = 0.1763
tan 50° = 1.1918
tan 70° = 2.7475
tan 10° - tan 50° tan 70° = 0.1763 + (1.1918 * 2.7475)
= 0.1763 + ( 3.27447)
= 3.45077
= 3.4508
tan 10° = 0.1763
tan 50° = 1.1918
tan 70° = 2.7475
tan 10° - tan 50° tan 70° = 0.1763 + (1.1918 * 2.7475)
= 0.1763 + ( 3.27447)
= 3.45077
= 3.4508
Answered by
3
tan(A + B) = (tan A + tan B) /1 − tan A tan B
tan(A − B) = (tan A − tan B)/ 1 + tan A tan B
consider A=60,B=10
A+B=70;A-B=50
thus we get the entire equation interms of tan(60) and tan(10)
tan(A+B)*tan(A-B)=((tan A)^2 --(tan B)^2)/1-(tan(A)*tan (B))^2
tan(A − B) = (tan A − tan B)/ 1 + tan A tan B
consider A=60,B=10
A+B=70;A-B=50
thus we get the entire equation interms of tan(60) and tan(10)
tan(A+B)*tan(A-B)=((tan A)^2 --(tan B)^2)/1-(tan(A)*tan (B))^2
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