Question

can someone answer it fast​

Answer 1

hint for your question

separate upper in two terms

(sin²titha + cos²titha) (sin²titha+cos²titha)

&

1= sin²titha+cos²titha

lower term get formula of

(a+b)²

then solve it .....by upper terms

hope u understand

Answer 2

Question

$\large{ \sf{ \dfrac{(sin {}^{4}x + {cos}^{4}x ) }{1 - 2 {sin}^{2}x \: {cos}^{2}x } = 1}}$

Answer

To Prove

$\large{ \sf{ \dfrac{(sin {}^{4}x + {cos}^{4}x ) }{1 - 2 {sin}^{2}x \: {cos}^{2}x } = 1}}$

LHS

$\large{ \sf{ \dfrac{(sin {}^{4}x + {cos}^{4}x ) }{1 - 2 {sin}^{2}x \: {cos}^{2}x }}} \\ \\ \large{ \longrightarrow \: \sf{ \frac{( {sin}^{2}x) {}^{2} + ({cos}^{2}x) {}^{2} }{1 - 2 {sin}^{2} x \: cos {}^{2} x} }}$

Since,

sin²∅ + cos²∅ = 1

$\large{ \longrightarrow \: \sf{ \dfrac{1}{( {sin}^{2}x + cos {}^{2}x - 2 {sin}^{2}xcos {}^{2}x) } }}$

$\large{ \longrightarrow \: \sf{ \dfrac{1}{1 - 0} }} \\ \\ \large{\longrightarrow \: \sf{1 } \longrightarrow \: RHS}$

Hence,Proved