2 years ago

Need to pass this semester so please help

Mathematics

1 answer:

Aleks04 [339]2 years ago

7 0

The correct answer is X= 14°

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Prove that :-<br><br> Cos3A-cosA/sin3A-sinA+cos2A-cos4A/sin4A-sin2A=sinA/cos2Acos3A

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$\dfrac{\cos^3A-\cos A}{\sin^3A-\sin A}+\dfrac{\cos^2A-\cos^4A}{\sin^4A-\sin^2A}=\dfrac{\sin A}{\cos^2A\cos^3A}\\\\L_s=\dfrac{\cos A(\cos^2A-1)}{\sin A(\sin^2A-1)}+\dfrac{\cos^2A(1-\cos^2A)}{\sin^2A(\sin^2A-1)}\\\\=\dfrac{\cos A(-\sin^2A)}{\sin A(-\cos^2A)}+\dfrac{\cos^2A\sin^2A}{\sin^2A(-\cos^2A)}\\\\=\dfrac{\sin A}{\cos A}-1=\tan A-1$

$R_s=\dfrac{\sin A}{\cos A\cos^4A}=\dfrac{\sin A}{\cos A}\cdot\dfrac{1}{\cos^4A}=\tan A\cdot\dfrac{1}{\cos^4A}=\dfrac{\tan A}{\cos^4A}$

$\boxed{L_s\neq R_s}$

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Step-by-step explanation:

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Answer C.

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