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Trigonometric ratios - class-IX
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$\sin ^{ 8 }{ \theta } -\cos ^{ 8 }{ \theta } -\left( \sin ^{ 2 }{ \theta } -\cos ^{ 2 }{ \theta } \right) \left( 1-\sin ^{ 2 }{ \theta } \right) $=0
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A
False
💡 Explanation:
$\sin^4 \theta+\cos^4 \theta=(\sin^2 \theta+\cos^2 \theta)^2-2\sin^2 \theta\cos^2 \theta=1-2\sin^2 \theta\cos^2 \theta$
$\sin^4 \theta-\cos^4 \theta=(\sin^2 \theta-\cos^2 \theta)(\sin^2 \theta+\cos^2 \theta)=\sin^2 \theta-\cos^2 \theta$
$\sin^8\theta-\cos^8\theta-(\sin^2 \theta-\cos^2 \theta)(1-\sin^2 \theta)=(\sin^4 \theta+\cos^4 \theta)(\sin^4 \theta-\cos^4 \theta)-(\sin^2 \theta-\cos^2 \theta)(1-\sin^2 \theta)$
$=(1-2\sin^2 \theta\cos^2 \theta)(\sin^2 \theta-\cos^2 \theta)-(\sin ^2 \theta-\cos^2\theta)(1-\sin^2 \theta)$
$=(\sin^2 \theta-\cos^2 \theta)(\sin^2 \theta)(1-2\cos^2 \theta)$
So the given relation is $\text{False}$