Question
Prove: $\sin ^6 A+3 \sin ^2 A \cos ^2 A=1-\cos ^6 A$
✓
Answer
Given- $\sin ^6 A+3 \sin ^2 A \cos ^2 A=1-\cos ^6 A$
Now, taking
$\sin ^6 A+\cos ^6 A=1-3 \sin ^2 A \cos ^2 A$
Taking $\ce{LHS}$
$=\sin ^6 A+\cos ^6 A=\left(\sin ^2 A\right)^3+\left(\cos ^2 A\right)^3$
$=\left(\sin ^2 A+\cos ^2 A\right)^3-3 \sin ^2 A \cos ^2 A\left(\sin ^2 A+\cos ^2 A\right)\left\{\because a^3+b^3=(a+b)^3-3 a b(a+b)\right\}$
$=(1)^3-3 \sin ^2 A \cos ^2 A(1)$
$=1-3 \sin ^2 A \cos ^2 A=\text { RHS }$
Now, taking
$\sin ^6 A+\cos ^6 A=1-3 \sin ^2 A \cos ^2 A$
Taking $\ce{LHS}$
$=\sin ^6 A+\cos ^6 A=\left(\sin ^2 A\right)^3+\left(\cos ^2 A\right)^3$
$=\left(\sin ^2 A+\cos ^2 A\right)^3-3 \sin ^2 A \cos ^2 A\left(\sin ^2 A+\cos ^2 A\right)\left\{\because a^3+b^3=(a+b)^3-3 a b(a+b)\right\}$
$=(1)^3-3 \sin ^2 A \cos ^2 A(1)$
$=1-3 \sin ^2 A \cos ^2 A=\text { RHS }$
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