Question

1 Prove that
$sin^{4}a - cos^{4} = 1 - 2cos^{2} a$


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Answer 1

Correct Question:

Prove that :

$\sf{{sin}^{4} \: A - {cos}^{4} \: A = 1 - 2{cos}^{2} \: A}$

Answer :-

Given to prove :

• $\sf{{sin}^{4} \: A - {cos}^{4} \: A = 1 - 2{cos}^{2} \: A}$

Solution :

L.H.S = $\sf{{sin}^{4} \: A - {cos}^{4} \: A}$

$\sf{\longrightarrow \: {sin}^{4} \: A - {cos}^{4} \: A}$

$\sf{\longrightarrow \: ({sin}^{2} \: A {)}^{2} - ({cos}^{2} \: A {)}^{2}}$

$\sf{\longrightarrow \: ({sin}^{2} \: A - {cos}^{2} \: A) \: ({sin}^{2} \: A + {cos}^{2} \: A)}$

$\sf{\longrightarrow \: ({sin}^{2} \: A - {cos}^{2} \: A) \: (1)} \: \: \: \: {}_{.. \: .. \: ..} \: \lgroup \sf{({sin}^{2} \: A + {cos}^{2} \: A) = 1} \rgroup$

$\sf{\longrightarrow \: {sin}^{2} \: A - {cos}^{2} \: A}$

$\sf{\longrightarrow \: (1 - {cos}^{2} \: A) - {cos}^{2} \: A} \: \: \: \: {}_{.. \: .. \: ..} \: \lgroup \sf{(1 - {cos}^{2} \: A) = {sin}^{2} \: A} \rgroup$

$\sf{\longrightarrow \: 1 - {cos}^{2} \: A - {cos}^{2} \: A}$

$\sf{\longrightarrow \: 1 - 2{cos}^{2} \: A = R.H.S}$

Hence, Proved!

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Things to remember :-

• $\sf{1 + {cos}^{2} \: A = {sin}^{2} \: A}$

• $\sf{{sin}^{2} \: A + {cos}^{2} = 1}$