Question

guys plz solve this.... .
prove it ​

Answer 1

To prove:-

${sec}^{4} \alpha (1 - {sin}^{4} \alpha ) - 2 {tan}^{2} \alpha = 1$

proof:-Take LHS

$= {sec}^{4} \alpha - \frac{ {sin}^{4} \alpha }{ {cos}^{4} \alpha } - 2 {tan}^{2} \alpha \\ = {sec}^{4} \alpha - {tan}^{4} \alpha - 2 {tan}^{2} \alpha \\ we \: know \: {sec}^{2} \alpha = 1 + {tan}^{2} \alpha \\ = ( {1 + {tan}^{2} \alpha })^{2} - {tan}^{4} \alpha - 2 {tan}^{2} \alpha \\ = 1 + {tan}^{4} \alpha + 2 {tan}^{2} \alpha - {tan}^{4} \alpha - 2 {tan}^{2} \alpha \\ cancelling \: out \: we \: get \\ </strong><strong>LH</strong><strong>S = 1 = </strong><strong>RHS</strong><strong>$

Hence proved