If cot theta =3/5 then what is the value of 6 tan theta-5cos theta

1.	If cot theta =3/5 then what is the value of 6 tan theta-5cos theta
Answer» $\blue{\bold{\underline{\underline{Answer:}}}}$ $\green{\tt{\therefore{6\:tan\:\theta-5\:cos\:\theta=7.42}}}$ $\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$ $\green{\underline \bold{<klux>GIVEN</klux> :}} \\ \tt: \implies cot \: \theta = \frac{3}{5} \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies 6 \: tan \: \theta - 5 \: cos \: \theta = ?$ • ACCORDING to given QUESTION : $\bold{As \: we \: know \: that} \\ \tt: \implies cot \: \theta = \frac{3}{5} \\ \\ \tt: \implies \frac{b}{p} = \frac{3}{5} \\ \\ \tt \circ \: base = 3 \\ \\ \tt \circ \: perpendicular = 5 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies h = \sqrt{ {p}^{2} + {b}^{2} } \\ \\\tt: \implies h = \sqrt{ {5}^{2} + {3}^{2} } \\ \\ \tt: \implies h = \sqrt{25 + 9} \\ \\ \tt: \implies h = \sqrt{34} \\ \\ \bold{for \: finding \: value} \\ \tt: \implies 6 \:tan \: \theta - 5 \: cos \: \theta \\ \\ \tt: \implies 6 \times \frac{p}{b} - 5 \times \frac{b}{h} \\ \\ \tt: \implies 6 \times \frac{5}{3 } - 5 \times \frac{3}{ \sqrt{34} } \\ \\ \tt: \implies 2 \times 5 - \frac{15}{ \sqrt{34} } \\ \\ \tt: \implies \frac{10 \sqrt{34} - 15}{ \sqrt{34} } \\ \\ \tt: \implies \frac{10 \times 5.83 - 15}{5.83} \\ \\ \tt: \implies \frac{58.3 - 15}{5.83} \\ \\ \green{\tt: \implies 7.42}$

Answer»

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{6\:tan\:\theta-5\:cos\:\theta=7.42}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{<klux>GIVEN</klux> :}} \\ \tt: \implies cot \: \theta = \frac{3}{5} \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies 6 \: tan \: \theta - 5 \: cos \: \theta = ?$

• ACCORDING to given QUESTION :

$\bold{As \: we \: know \: that} \\ \tt: \implies cot \: \theta = \frac{3}{5} \\ \\ \tt: \implies \frac{b}{p} = \frac{3}{5} \\ \\ \tt \circ \: base = 3 \\ \\ \tt \circ \: perpendicular = 5 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies h = \sqrt{ {p}^{2} + {b}^{2} } \\ \\\tt: \implies h = \sqrt{ {5}^{2} + {3}^{2} } \\ \\ \tt: \implies h = \sqrt{25 + 9} \\ \\ \tt: \implies h = \sqrt{34} \\ \\ \bold{for \: finding \: value} \\ \tt: \implies 6 \:tan \: \theta - 5 \: cos \: \theta \\ \\ \tt: \implies 6 \times \frac{p}{b} - 5 \times \frac{b}{h} \\ \\ \tt: \implies 6 \times \frac{5}{3 } - 5 \times \frac{3}{ \sqrt{34} } \\ \\ \tt: \implies 2 \times 5 - \frac{15}{ \sqrt{34} } \\ \\ \tt: \implies \frac{10 \sqrt{34} - 15}{ \sqrt{34} } \\ \\ \tt: \implies \frac{10 \times 5.83 - 15}{5.83} \\ \\ \tt: \implies \frac{58.3 - 15}{5.83} \\ \\ \green{\tt: \implies 7.42}$