Question

Plz solve question no 12

Answer 1

$\frac{( { \sin }^{4} \theta +{\cos }^{4} \theta) }{1 - 2{ \sin}^{2} \theta { \cos}^{2} \theta} = 1\\ { \sin}^{2} \theta \: + { \cos }^{2} \theta = 1(identity)$
Squaring both sides,
$( {{ \sin}^{2} \theta \: + { \cos }^{2} \theta})^{2} = {1}^{2} \\ { \sin}^{4} \theta + { \cos }^{4} \theta + 2 { \sin}^{2} \theta { \cos}^{2} \theta = 1 \\ { \sin}^{4} \theta + { \cos }^{4} \theta = 1 - 2 { \sin}^{2} \theta { \cos}^{2} \theta \\ { \sin}^{4} \theta + { \cos }^{4} \theta = (1 - 2 { \sin}^{2} \theta { \cos}^{2} \theta) \times 1 \\ \frac{({ \sin}^{4} \theta + { \cos }^{4} \theta)}{1 - 2 { \sin}^{2} \theta { \cos}^{2} \theta} = 1$
Hence, proved.

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Answer 2

(sin^4+cos^4)
_____________. =1
1-2sin^2cos^2

(sin^2+cos^2)^2-2sin^2cos^2
_______________________=1
1-2sin^2cos^2

1-2sin^2cos^2
____________ =1. ( sin^2+ cos^2 = 1 )
1-2sin^2cos^2

1=1 (PROVED)

Thank You!