Question

prove that sec^2Ø-(sin^2ø-2sin^4ø/2cos^4-cos^2ø)=1​

Answer 1

$\large\sf\underline{ \underline{ \red{solution : }}}$

$\\ \implies \: \sf{ {sec}^{2}\theta - \frac{ {sin}^{2}\theta \: - 2 {sin}^{4}\theta \: }{2 {cos}^{4}\theta \: - {cos}^{2} \theta \: } = 1} \\ \\ \\ \implies \: \sf{ {sec}^{2}\theta \: - \frac{ {sin}^{2}\theta \: (1 - 2 {sin}^{2}\theta \: ) }{ {cos}^{2} \theta \: (2 {cos}^{2} \theta \: - 1)} } \\ \\ \\ \implies \: \sf{ {sec}^{2}\theta \: - \frac{ {sin}^{2} \theta \: (1 - {sin}^{2} \theta \: - {sin}^{2}\theta \: ) }{ {cos}^{2}\theta \: (2 {cos }^{2}\theta \: - ( {sin}^{2}\theta \: + {cos}^{2}\theta \: ))} } \\ \\ \\ \implies \: \sf{ {sec}^{2}\theta \: - \frac{ {sin}^{2}\theta \: ( {cos}^{2}\theta \: - {sin}^{2}\theta \: )}{ {cos}^{2}\theta \: (2 {cos}^{2}\theta \: - {sin}^{2}\theta \: - {cos}^{2} \theta \: )} } \\ \\ \\ \implies \: \sf{ {sec}^{2} \theta \: - \frac{ {sin}^{2} \theta \: ( \cancel{ {cos}^{2}\theta \: - {sin}^{2} \theta \: )}}{ {cos}^{2}\theta \: ( \cancel{ {cos}^{2} \theta \: - {sin}^{2}\theta \: )} }} \\ \\ \\ \implies \: \sf{ \frac{1}{ {cos}^{2}\theta \: } - \frac{ {sin}^{2} \theta \: }{ {cos}^{2} \theta \: } } \\ \\ \\ \implies \: \sf{ \frac{1 - {sin}^{2}\theta \: }{ {cos}^{2}\theta \: } } \\ \\ \\ \implies \: \sf{ \cancel{ \frac{ {cos}^{2} \theta \: }{ {cos}^{2} \theta \: } }} \\ \\ \\ \implies \: \sf{ 1 \: \: \: \: \: \: \: \: \: \: \: (proved)}$

Important Identities :-

$\\ \\ \sf (1) \: \: {sin}^{2} \theta + {cos}^{2} \theta = 1 \\ \\ \\ \sf (2) \: \: 1 + {cot}^{2} \theta = {cosec}^{2} \theta \\ \\ \\ \sf(3) \: \: 1 + {tan}^{2} \theta = {sec}^{2} \theta$