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sin⁴A - cos⁴A = 2sin²A - …

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sin⁴A - cos⁴A = 2sin²A - 1

Posted by Mr Harry 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

We have,
L.H.S = sin⁴A - cos⁴A
{tex}\Rightarrow{/tex} L.H.S = (sin²A)² - (cos²A)²
{tex}\Rightarrow{/tex} L.H.S = (sin²A + cos²A) (sin²A - cos²A) {tex}\left[\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right]{/tex}
{tex}\Rightarrow{/tex} L.H.S = sin²A - cos² A [ {tex}\therefore{/tex} sin² A + cos² A = 1]
{tex}\Rightarrow{/tex} L.H.S = (1 - cos²A)- cos² A = 1 - 2cos² A = 1-2(1-sin²A) = 2sin²A-1 = R.H.S

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