Question

if 4tanA=3 then evaluate 4sinA-cosA+1/4sinA+cosA-1

Answer 1

4sinA-cosA+1/4sinA+cosA-1 = 13/11

Answer 2

GIVEN:
4tanA=3
tanA = ¾ = P/B
In a ∆ ABC right angled at B.
Base (AB) = 4 , Perpendicular( BC) = 3
AC² = AB² + BC²
[By Pythagoras theorem]
AC² = 4² + 3²
AC² = 16 + 9 = 25
AC= √25 = 5
The trigonometric ratios of ∠A,
Base (AB)= 4, perpendicular (BC)= 3 , hypotenuse(AC) = 5
sin A = perpendicular /hypotenuse = BC /AC sin A = ⅗
Cos A = Base/hypotenuse = ⅘
4sinA-cosA+1/4sinA+cosA-1
= 4×3/5 - ⅘ +1 / 4×3/5 + ⅘ -1
= 12/5 - 4+5/5 / 12/5 + (4 -5/5)
= 12/5 +⅕ / 12/5 -⅕
= 13/5 / 11/5
= 13/5 × 5/11 = 13/11
4sinA-cosA+1/4sinA+cosA-1 = 13/11.

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