Question

If sin A = 4/5 find cos A + tan A.​

Answer 1

as we know

$\sin(e) = \frac{perpendicular}{hypotenuse} = \frac{4x}{5x} = \frac{4}{5}$

so , by Pythagoras theorem

H² = B² + P²

${(5x)}^{2} = {b}^{2} + {4x}^{2} \\ \sqrt{ 25{ x }^{2} - 16 {x}^{2} } = base \\ \sqrt{9 {x}^{2} } = 3x = base$

now ,

$\cos(e) = \frac{base}{hypotenuse} = \frac{3x}{5x} = \frac{3}{5} \\ \tan(e) = \frac{perpendicular}{base} = \frac{4x}{3x} = \frac{4}{3}$

$\cos(e) + \tan(e) \\ = \frac{3}{5} + \frac{4}{3} \\ = \frac{9 + 20}{15} \\ = \frac{29}{15}$

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now,