Question

if 4 tan0=3 ecaluate (4 sin 0-cos0+1÷4sin 0+ cos0-1)
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Answer 1

4 tanA=3

tanA = 3/4

by Pythagoras theorem

H² = P²+B²

H² = 9k² +16k² = 25k²

H= 5k

sinA = 3k/5k = 3/5

cosA = 4k/5k = 4/5

(4 sin A-cosA+1 / 4sin A+ cosA-1)

(4×3/5 - 4/5+1 / 4× 3/5 + 4/5 -1 )

12/5 -4/5 +1 / 12/5 +4/5 -1

8/5 +1 / 16/5 -1

13/5 / 11/5

13/11

Answer 2

Answer:

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