Question

If 5tanθ=4, then value of (5sinθ -4cosθ)/(5sinθ +4cosθ) is:​

Answer 1

${\huge{\underline{\small{\mathbb{\pink{REFER \ TO \ THE \ ATTACHMENT}}}}}}$

Answer 2

$\large{\underline{\underline{ \bf{⎆Question:-}}}}$

• If 5tanθ = 4, then value of (5sinθ - 4cosθ)/(5sinθ +4cosθ) is :

$\large{\underline{\underline{ \bf{☂ Solution:-}}}}$

$\bold{ \large \underline{ \mathbb{\underline{GIVEN : }}}}$

$\longmapsto\tt{5 \tanθ =4 }$

$\longmapsto\tt{ \tanθ = \: \frac{4}{5} }$

$\longmapsto\tt{ \tanθ = \: \frac{p}{b} }$

$\longmapsto\tt{ \frac{4}{5} = \: \frac{p}{b} }$

$\longmapsto\tt{p =4 \: and \: b = 5 }$

$\small\underline\mathtt\pink{✰From\: Pythagoras'\: Theorem}$

$\longmapsto\tt{ {h}^{2} = {p}^{2} + {b}^{2} }$

$\longmapsto\tt{ {h}^{2} = {4}^{2} + {5}^{2} }$

$\longmapsto\tt{ {h}^{2} = 16 + 25 }$

$\longmapsto\tt{ {h} = \sqrt{41} }$

$\small\underline\mathtt\pink{✰Thus \: we \: have \: to \: find \: the \: value \: of \: sin θ \: and \: cos θ .}$

$\longmapsto\tt{ \sin θ = \: \frac{p}{h} \: and \: cos θ = \frac{b}{h} }$

$\longmapsto\tt{ \sin θ = \: \frac{4}{ \sqrt{41} } \: and \: cos θ = \frac{5}{ \sqrt{41} } }$

$\small\underline\mathtt\pink{✰Therefore \: - }$

$\longmapsto\tt{ \frac{5 sin θ - 4 \cos θ }{5 sin θ + 4 \cos θ } }$

$\longmapsto\tt{ \frac{5 \times \frac{4}{ \sqrt{41} } - 4 \times \frac{5}{ \sqrt{41} } }{5 \times \frac{4}{ \sqrt{41} } + 4 \times \frac{5}{ \sqrt{41} } } }$

$\longmapsto\tt{ \frac{ \frac{20}{ \sqrt{41} } - \frac{20}{ \sqrt{41} } }{ \frac{20}{ \sqrt{41} } + \frac{20}{ \sqrt{41} } } }$

$\longmapsto\tt{ \frac{0}{0} }$

$\small\underline\mathtt\pink{✰Alternative \: \: method \: - }$

$\tt{Divide\: numerator\:and\: denominator\:by\: cos θ}$

$\longmapsto\tt{ \: \frac{5tan θ - 4}{5tan θ + 4} }$

$\longmapsto\tt{ \: \frac{5 \times \frac{4}{5} - 4}{5 \times \frac{4}{5} + 4} }$

$\longmapsto\tt{ \: \frac{0}{8}\:or\:0 }$

• Hope it helps.