Question

sin⁴x-cos⁴x/sin²x-cos²x= 1​

Answer 1

${\bold{\huge{\underline{\blue{\sf{Question:}}}}}}$

${\sf{Prove \: \: \frac {(sin⁴x - cos⁴x)}{(sin²x - cos²x )}= \: 1}}$

${\bold{\huge{\underline{\orange{\sf{Formulas:}}}}}}$

${\boxed{\sf{sin²x + cos²x = 1}}}$

${\boxed{\sf{a² - b² = (a + b)(a - b)}}}$

${\bold{\huge{\underline{\red{\sf{Solution:}}}}}}$

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

${\boxed{\sf{★ \: Hence \: proved \: ★}}}$

Answer 2

Answer:

sin⁴x-cos⁴x/sin²x-cos²x= 1

Answer:-

LHS:-

=sin⁴x-cos⁴x/sin²x-cos²x

(sin²x)²-(cos²x)²

=__________________

sin²x-cos²x

(sin²x+cos²x)(sin²x-cos²x)

=____________________

sin²x-cos²x

The numerator and denominator get cancelled,

And sin²x+cos²x will remain....

As per theorem,

sin²A+cos²A=1

Hope it will helps you...............

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