Question

Please solve this quickly

Answer 1

$\mathsf{Given :\;\dfrac{2.sin\theta.cos\theta}{cos^2\theta - sin^2\theta}}$

The Above Fraction can be written as :

$\mathsf{\implies \dfrac{2}{\dfrac{cos^2\theta - sin^2\theta}{sin\theta.cos\theta}}}$

$\mathsf{\implies \dfrac{2}{\dfrac{cos^2\theta}{sin\theta.cos\theta} - \dfrac{sin^2\theta}{sin\theta.cos\theta}}}$

$\mathsf{\implies \dfrac{2}{\dfrac{cos\theta}{sin\theta} - \dfrac{sin\theta}{cos\theta}}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{\dfrac{cos\theta}{sin\theta} = cot\theta}}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{\dfrac{sin\theta}{cos\theta} = tan\theta}}}$

$\mathsf{\implies \dfrac{2}{cot\theta - tan\theta}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{cot\theta = \dfrac{1}{tan\theta}}}}$

$\mathsf{\implies \dfrac{2}{\dfrac{1}{tan\theta} - tan\theta}}$

$\mathsf{\implies \dfrac{2}{\dfrac{1 - tan^2\theta}{tan\theta}}}$

$\mathsf{\implies \dfrac{2.tan\theta}{1 - tan^2\theta}}$

$\mathsf{Given :\;tan\theta = \dfrac{12}{13}}$

$\mathsf{\implies \dfrac{2\bigg(\dfrac{12}{13}\bigg)}{1 - \bigg(\dfrac{12}{13}\bigg)^2}}$

$\mathsf{\implies \dfrac{\bigg(\dfrac{24}{13}\bigg)}{1 - \bigg(\dfrac{144}{169}\bigg)}}$

$\mathsf{\implies \dfrac{\bigg(\dfrac{24}{13}\bigg)}{\bigg(\dfrac{169 - 144}{169}\bigg)}}$

$\mathsf{\implies \dfrac{\bigg(\dfrac{24}{13}\bigg)}{\bigg(\dfrac{25}{169}\bigg)}}$

$\mathsf{\implies \bigg(\dfrac{24}{13}\times \dfrac{169}{25} \bigg)}$

$\mathsf{\implies \bigg({24}\times \dfrac{13}{25} \bigg)}$

$\mathsf{\implies \dfrac{312}{25}}$