Question

prove:
$\sec {}^{4} \alpha + { \tan {}^{4} \alpha } = 1 + 2 \sec {}^{2} \alpha \tan{}^{2} \alpha$
​

Answer 1

$\bold{\pink{To \: prove}}\longrightarrow \\ {sec}^{4} \alpha + {tan}^{4} \alpha = 1 + 2 {sec}^{2} \alpha \: {tan}^{2} \alpha$

$\bold{\green{Concept \: used}}\longrightarrow \\ 1)1 + {tan}^{2} \alpha = {sec}^{2} \alpha \\ = > 1 = {sec}^{2} \alpha - {tan}^{2} \alpha \\ 2) {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy$

$\bold{\blue{Solution}}\longrightarrow \\\pink{LHS} \\ = {sec}^{4} \alpha + {tan}^{4} \alpha \\ = {(sec}^{2} \alpha) ^{2} + { ({tan}^{2} \alpha )}^{2} \\adding \: and \: subtracting \: {sec}^{2} \alpha {tan}^{2} \alpha \\ = { ({sec}^{2} \alpha) }^{2} + {( {tan}^{2} \alpha ) }^{2} - 2 {sec}^{2} \alpha {tan}^{2} \alpha + 2 {sec}^{2} \alpha {tan}^{2} \alpha \\ = {( {sec}^{2} \alpha - {tan}^{2} \alpha ) }^{2} + 2 {sec}^{2} \alpha \: {tan}^{2} \alpha \\ now \: , {sec}^{2} \alpha - {tan}^{2} \alpha = 1 \:, using \: it \: \\ = {(1)}^{2} + 2 {sec}^{2} \alpha {tan}^{2} \alpha \\ = 1 + 2 {sec}^{2} \alpha {tan}^{2} \alpha \\ =\pink{ RHS}$

$\bold{\red{Additional \: information}}\longrightarrow \\1) {sin}^{2} \alpha + {cos}^{2} \alpha = 1 \\ 2)1 + {cot}^{2} \alpha = {cosec}^{2} \alpha \\ 3)sin(90 ^{0} - \alpha ) = cos \alpha \\ 4) \cos(90 ^{0} - \alpha ) = sin \alpha$