Question

If $sec\theta =\frac{5}{4},$ find the value of $\frac{sin\theta -2cos\theta }{tan\theta -cot\theta }$

Answer 1

Given,

$secA=\frac{5}{4}$

Therefore, $cosA=\frac{1}{\sec \theta }=\frac{4}{5}$

We know that,

$cosA=\frac{adjacentSide}{Hypotenuse}$

From Pythagoras theorem,

$(Hypotenuse{)}^2=(oppositeSide{)}^2+(adjacentSide{)}^2$

$5^2=(oppositeSide{)}^2+4^2$

$(oppositeSide{)}^2=25-16=9$

$\left(oppositeSide\right)=3$

$sinA=\frac{OppositeSide}{Hypotenuse}=\frac{3}{5}$

$tanA=\frac{OppositeSide}{AdjacentSide}=\frac{3}{4}$

$\cot \theta =\frac{1}{\tan \theta }=\frac{4}{3}$

Therefore,

$\frac{\sin \theta -2\cos \theta }{\tan \theta -\cot \theta }=\frac{\frac{3}{5}-2\left(\frac{4}{5}\right)}{\frac{3}{4}-\frac{4}{3}}$

$=\frac{\left(\frac{-5}{5}\right)}{\left(\frac{-7}{12}\right)}$

$=\frac{12}{7}$