if secA+tanA=5 find the quadrant in which A lies,find sinA.
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We know that, If secA+tanA=p, then secA-tanA=1/p
Given,
secA+tanA=5
Then secA-tanA=1/5
Case 1:
By adding 1 and 2 we get,
2secA=5+1/5 (since, +tanA and -tanA gets cancelled)
2secA = 25+1/5 (LCM)
secA = 26/5 × 1/2
secA = 13/5
secA will be positive on Q1 and Q4 ------- 1`
Case 2:
By subtracting 1 and 2
2tanA = 5-1/5 (since, +secA and -secA gets cancelled)
2tanA = 25-1/5 (LCM)
tanA = 24/5 × 1/2
tanA = 12/5
tanA will be positive on Q1 and Q3 ------- 2
from 1 and 2 markings, Q1 is common
so, A belongs to Q1
From calculation, we get tanA = 12/5 = opp side/ adj side
Then hypotenuse = 13 (by pythagoras theorem)
Then, sinA = opp side/hypotenuse = 12/13
Therefore, sinA = 12/13
Given,
secA+tanA=5
Then secA-tanA=1/5
Case 1:
By adding 1 and 2 we get,
2secA=5+1/5 (since, +tanA and -tanA gets cancelled)
2secA = 25+1/5 (LCM)
secA = 26/5 × 1/2
secA = 13/5
secA will be positive on Q1 and Q4 ------- 1`
Case 2:
By subtracting 1 and 2
2tanA = 5-1/5 (since, +secA and -secA gets cancelled)
2tanA = 25-1/5 (LCM)
tanA = 24/5 × 1/2
tanA = 12/5
tanA will be positive on Q1 and Q3 ------- 2
from 1 and 2 markings, Q1 is common
so, A belongs to Q1
From calculation, we get tanA = 12/5 = opp side/ adj side
Then hypotenuse = 13 (by pythagoras theorem)
Then, sinA = opp side/hypotenuse = 12/13
Therefore, sinA = 12/13
TPS:
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