Question

If $sin\theta =\frac{12}{13},$ find the value of $\frac{si{n}^2\theta -co{s}^2\theta }{2sin\theta cos\theta }\times \frac{1}{ta{n}^2\theta }$

Answer 1

Given,

$sinA=\frac{12}{13}$

We know that,

$sinA=\frac{oppositeSide}{Hypotenuse}$

From Pythagoras theorem,

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

${13}^2={12}^2+(adjacentSide{)}^2$

$(adjacentSide{)}^2=169-144=25$

$\left(adjacentSide\right)=5$

$cosA=\frac{AdjacentSide}{Hypotenuse}=\frac{5}{13}$

$tanA=\frac{OppositeSide}{AdjacentSide}=\frac{12}{5}$

Therefore,

$\frac{{\sin }^2\theta -{\cos }^2\theta }{2\sin \theta \cos \theta }\times \frac{1}{{\tan }^2\theta }$

$=\frac{{\left(\frac{12}{13}\right)}^2-{\left(\frac{5}{13}\right)}^2}{2\left(\frac{12}{13}\right)\left(\frac{5}{13}\right)}\times \frac{1}{{\left(\frac{12}{5}\right)}^2}$

$=\frac{\left(\frac{144}{169}\right)-\left(\frac{25}{169}\right)}{2\left(\frac{12}{13}\right)\left(\frac{5}{13}\right)}\times \frac{25}{144}$

$=\frac{\left(\frac{144-25}{169}\right)}{\left(\frac{120}{169}\right)}\times \frac{25}{144}$

$=\frac{119}{120}\times \frac{25}{144}$

$=\frac{119}{24}\times \frac{5}{144}$

$=\frac{595}{3456}$