Question

3 tan thita =4 then find the value of 4 cos thita-sin thita/2 cos thita+sin thita

Answer 1

SOLUTION :

${\underline{\bf{Given\::}}}$

3 tan Ф = 4.

${\underline{\bf{Explanation\::}}}$

$\bigstar$ Firstly, attach a figure :

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\linethickness{0.4mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{B}}\put(10.6,1){\large\sf{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(7.2,2){\large\sf{4 \:cm}}\put(8.8,0.7){\large\sf{3\:cm}}\put(9.4,1.9){\large\sf{}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

We know that;

$\boxed{\sf{tan\theta=\frac{Perpendicular}{Base} }}}}$

$\longrightarrow\sf{3\:tan\:\theta=4}\\\\\longrightarrow\sf{tan\theta=4/3}$

$\therefore \bf{\dfrac{AB}{BC} =\dfrac{4}{3} }$

$\bigstar$ Secondly,using pythagoras theorem, we get hypotenuse :

$\longrightarrow\sf{(Hypotenuse)^{2} =(base)^{2} +(perpendicular)^{2} }\\\\\longrightarrow\sf{(AC)^{2} =(BC)^{2} +(AB)^{2} }\\\\\longrightarrow\sf{(AC)^{2} =(3)^{2} +(4)^{2} }\\\\\longrightarrow\sf{(AC)^{2} =9+16}\\\\\longrightarrow\sf{(AC)^{2}=25}\\\\\longrightarrow\sf{AC=\sqrt{25} }\\\\\longrightarrow\bf{AC=5}$

$\bigstar$ Thirdly, substituting the value in given :

$\longrightarrow\sf{\dfrac{4cos\theta-sin\theta }{2cos \theta + sin\theta} }\\\\\\\longrightarrow\sf{\frac{4\times \dfrac{Base}{Hypotenuse}-\dfrac{Perpendicular}{Hypotenuse} }{2\times\dfrac{Base}{Hypotenuse}+\dfrac{Perpendicular}{Hypotenuse} } }\\\\\\\longrightarrow\sf{\dfrac{4\times \dfrac{3}{5} -\dfrac{4}{5} }{2\times \dfrac{3}{5} + \dfrac{4}{5} } }\\\\\\\longrightarrow\sf{\dfrac{\dfrac{12}{5} -\dfrac{4}{5} }{\dfrac{6}{5} + \dfrac{4}{5} } }$

$\longrightarrow\sf{\dfrac{\dfrac{12-4}{5} }{\dfrac{6+4}{5} } }\\\\\\\longrightarrow\sf{\dfrac{\dfrac{8}{5} }{\dfrac{10}{5} } }\\\\\\\longrightarrow\sf{\dfrac{8}{\cancel{5}} \times \dfrac{\cancel{5}}{10} }\\\\\\\longrightarrow\sf{\cancel{\dfrac{8}{10} }}\\\\\\\longrightarrow\bf{\dfrac{4}{5} }$

Thus;

The value will be 4/5 .