If tan- - 47- then 7 sin- - 3cos-7sin- - 3cos- -

The correct option is C 17 7sinθ 3cosθ7sinθ + 3cosθ = 7 tanθ 37 tanθ + 3 (∵ dividing numerator and denominator by cosθ) = 7×47 37 ×4 ...

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Standard X

Mathematics

Trigonometric Identities

If tanθ = 4...

Question

If $tan θ = \frac{4}{7}$ , then $\frac{7 sin θ - 3 cos θ}{7 sin θ + 3 cos θ} =$ _____

\frac{1}{7}

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\frac{5}{7}

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\frac{3}{7}

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\frac{5}{14}

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Solution

The correct option is C $\frac{1}{7}$
$\frac{7 sin θ - 3 cos θ}{7 sin θ + 3 cos θ} = \frac{7 tan θ - 3}{7 tan θ + 3} (∵$ dividing numerator and denominator by $cos θ)$
$= \frac{7 \times \frac{4}{7} - 3}{7 \times \frac{4}{7} + 3} = \frac{1}{7}$ .

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Similar questions

If $7 tan θ = 4$ , then $\frac{(7 sin θ - 3 cos θ)}{(7 sin θ + 3 cos θ)} = ?$
(a) $\frac{1}{7}$
(b) $\frac{5}{7}$
(c) $\frac{3}{7}$
(d) $\frac{5}{14}$

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\frac{7}{r} = 5 + 3 cos θ + 4 sin θ

represents

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If tanθ=47, then 7sinθ−3cosθ7sinθ+3cosθ=_____

The correct option is C 177sinθ−3cosθ7sinθ+3cosθ=7tanθ−37tanθ+3(∵ dividing numerator and denominator by cosθ)=7×47−37×47+3=17.

If $tan θ = \frac{4}{7}$ , then $\frac{7 sin θ - 3 cos θ}{7 sin θ + 3 cos θ} =$ _____

The correct option is C $\frac{1}{7}$
$\frac{7 sin θ - 3 cos θ}{7 sin θ + 3 cos θ} = \frac{7 tan θ - 3}{7 tan θ + 3} (∵$ dividing numerator and denominator by $cos θ)$
$= \frac{7 \times \frac{4}{7} - 3}{7 \times \frac{4}{7} + 3} = \frac{1}{7}$ .