SinA=4/5 find value (3sin A-cosA)/(4cosecA 3tanA)
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SinA=4/5=P/H
hence P=4 H=5
By pythagoras theorem
H^2=B^2+P^2
25=B^2+16
B^2=9
B=3
CosA=B/H=3/5
CosecA=5/4
tanA=4/3
Hence,
=(3×4/5-3/5) (4×5/4×3×4/3)
=(12/5-3/5)(20/4×12/3)
=(9/5)(120/12)
=9/5×10
=90/5
=18ans...
hope it helps u...
hence P=4 H=5
By pythagoras theorem
H^2=B^2+P^2
25=B^2+16
B^2=9
B=3
CosA=B/H=3/5
CosecA=5/4
tanA=4/3
Hence,
=(3×4/5-3/5) (4×5/4×3×4/3)
=(12/5-3/5)(20/4×12/3)
=(9/5)(120/12)
=9/5×10
=90/5
=18ans...
hope it helps u...
Answered by
0
sinA = 4/5 &cosA = 3/5 & tanA = 4/3
cosecA = 5/4
(3sinA-cosA)/(4cosecA×3tanA) = 3/25
cosecA = 5/4
(3sinA-cosA)/(4cosecA×3tanA) = 3/25
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