Question

if sin^4a-cos^4a=b^4 then sin^2a-cos^2a is what?

Answer 1

Answer:

sin ^4a +- cos ^ 4 a =b^4 sin ^2 a - cos ^ 2a

Answer 2

Question :-

$\rm \: if \: { \sin}^{4} a - { \cos}^{4} a = {b}^{4} \: then \: what \: is \: the \: value \\ \rm \: of \: \: { \sin}^{2} a - { \cos}^{2} a \: .$

Answer :-

$\to \boxed{ \rm \: { \sin}^{2} a - { \cos}^{2} a = {b}^{4} } \:$

To Find :-

$\mapsto \rm { \sin}^{2} a - { \cos}^{2} a$

Used Identity :-

$\boxed{\to} \: \rm {x}^{2} - {y}^{2} = (x + y)(x - y) \\ \\ \boxed{ \to} \: \: { \sin}^{2} \theta + { \cos}^{2} \theta = 1$

Solution :-

According to the question,

$\to \rm{ \sin}^{4} a - { \cos}^{4} a = {b}^{4} \: \: \\ \\ \bf \: we \: can \: write \: it \: \\ \\ \to \rm { \{ { \sin}^{2}a \} }^{2} - { { \{ \cos}^{2}a \} }^{2} = {b}^{4} \\ \: \: \: \: \: \: \: \: \boxed{ \because \bf {x}^{2} - {y}^{2} = (x - y)(x + y)} \\ \therefore \\ \to \rm \big \{ { \sin}^{2} a + { \cos}^{2} a \big \} \big \{ { \sin}^{2} a - { \cos}^{2} a \big \} = {b}^{4} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \: { \sin}^{2} \theta + { \cos}^{2} \theta = 1} \\ \therefore \: \\ \to \boxed{ \rm \: { \sin}^{2} a - { \cos}^{2} a = {b}^{4} }$

Some Important Formula :-

Trigonometric identities:-

$\mapsto \: \sin 2\theta = 2 \sin \theta \: \cos \theta \\ \\ \mapsto \cos2\theta = 1 - 2 \sin^{2} \theta \\ \\ \mapsto \cos 2\theta = 2\cos^{2}\theta -1 \\ \\ \mapsto \cos2\theta = \cos^{2} \theta - \sin^{2} \theta \\ \\ \mapsto \sec^{2} \theta - \tan^{2} \theta = 1 \\ \\ \mapsto \csc^{2} \theta - \cot^{2} \theta = 1$