Question

1-tan2theta/cot2theta-1=tan2theta

Answer 1

$\huge{\bold{ Question:-}}$

To prove

$\frac{1 - {tan}^{2} \theta}{ {cot}^{2} \theta - 1 } = {tan}^{2} \theta \\ \\ {\purple{\boxed{\large{\bold{trigonometry \: identies \: used}}}}} \\ \\ \frac{ {sin}^{2} \theta }{ {cos}^{2} \theta } = {tan}^{2} \theta \\ \\ \frac{ {cos }^{2} \theta}{ {sin}^{2} \theta } = {cot}^{2} \theta \\ \\ \mid{\green{\underline{\red{\mathbb{Taking \: LHS}}}}} \\ \\ \frac{1 - {tan}^{2} \theta}{ {cot}^{2} \theta - 1} \\ \\ \implies \frac{1 - \frac{ {sin}^{2} \theta }{ {cos}^{2} \theta} }{ \frac{ {cos}^{2} \theta}{ {sin}^{2} \theta} - 1} \\ \\ \implies \: \frac{ \frac{ {cos}^{2} \theta - {sin}^{2} \theta }{ {cos}^{2} \theta } }{ \frac{ {cos}^{2} \theta - {sin}^{2} \theta}{ {sin}^{2} \theta } } \\ \\ \implies \frac{ {cos}^{2} \theta - {sin}^{2} \theta}{ {cos}^{2} \theta } \times \frac{ {sin}^{2} \theta }{ {cos}^{2} \theta - {sin}^{2} \theta } \\ \\ \implies \frac{ {sin}^{2} \theta}{ {cos}^{2} \theta} \\ \\ \implies {tan}^{2} \theta \: = RHS \\ \\ {\blue{\boxed{\large{\bold{LHS \: = RHS}}}}} \\ \\ \huge{\bold{Hence \: proved}}$

Hope its helpful ❤️