If secA = x + 1/4 x then secA + tanA = ?
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Answers
Step-by-step explanation:
GIVEN:
sec A = x+(1/4x)
TO PROVE:
sec A+tan A = 2x or (1/2 x)
PROOF:
sec A = x+(1/4x) {equation 1 }
Squaring both sides
= sec² A = [x+(1/4x)]²
= sec² A = (x)² + [(1/4 x)]² + 2.x.(1/4x) {because (a+b)² = (a)² + (b)² + 2ab }
= sec² A = x² + 1/16x² + 1/2
Since sec² A = 1 + tan² A,
So,
1+ tan² A = x² + 1/16x² + 1/2
= tan² A = x² + 1/16x² + 1/2 - 1
= tan² A = x² + 1/16x² + (1-2/2)
= tan² A = x² + 1/16x² - 1/2
= (tan A)² = [x-(1/4x)]² {because: (a)² + (b)² - 2ab = (a-b)² }
=> tan A = ± [x-(1/4x)] {equation 2 }
Adding equation 1 and equation 2:
= sec A + tan A = [x+(1/4x)] ± [ x-(1/4x)]
IF IT IS +[ x-(1/4x)] THEN,
sec A + tan A = (x+1/4x)+(x-1/4x)
sec A + tan A = x + 1/4x + x - 1/4x
sec A + tan A = 2x {equation I }
IF IT IS -[ x-(1/4x)]
sec A + tan A = (x+1/4x) - (x-1/4x)
sec A + tan A = x+1/4x - x + 1/4x
sec A + tan A = 2/4x
sec A + tan A = 1/2x {equation II }
From equation I and equation II,
sec A + tan A = 2x or 1/2x
HENCE, PROVED