Question

If 4tan0=3.
Evaluate: (4sino-coso+1)/4sino+coso-1)​

Answer 1

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This question was asked in [Board 2019]

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Answer 2

Step-by-step explanation:

$GIVEN: \\ </p><p> = > 4tanA=3</p><p>tanA = \frac{3}{4} = \frac{p}{b} \\ </p><p> \: In \: a \: ABC \: right \: angled \: at B. \\ Base \: (AB) = 4 , \\ Perpendicular( BC) = 3 \: \\ </p><p> = > {AC}^{2} = {AB}^{2} + {BC}^{2} . \\ = > {AC}^{2} = 16 + 9 = 25 \\ </p><p> = > AC= \sqrt{25} = 5 \: </p><p>The \: trigonometric \: \\ ratios \: of. \\ ac = 5 \\ sin A = perpendicular /hypotenuse = \frac{BC }{ \frac{3}{5} } \\ = > Cos A = Base/hypotenuse = \frac{4}{5} \\ = > 4× \frac{3}{5} - \frac{4}{5} + \frac{1}{4} × \frac{3}{5} + \frac{4}{5} -1 \\ = > 125 - (4 - \frac{5}{5} ) / \frac{12}{5} + (4 - \frac{5}{5} ) \\ = > </p><p>\frac{13}{5} / \frac{11}{5} </p><p>= \frac{13}{5} × \frac{5}{11} = \frac{13}{11} \\ = > \frac{12}{5} + \frac{1}{5} / \frac{12}{5} - \frac{1}{5} \\ = > \frac{13}{5} / \frac{11}{5} </p><p>= \frac{13}{5} × \frac{5}{11} = \frac{13}{11} \\ = > 4sinA-cosA+1/4sinA+cosA-1 = 13/11. \\ hope \: it \: helps \: you</p><p>$