The equation frac{sin^3 theta - cos^3 theta}{ | vedclass

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Question

(A) Given the expression: $\frac{\sin^3 	heta - \cos^3 	heta}{\sin 	heta - \cos 	heta} - \frac{\cos 	heta}{\sqrt{1 + \cot^2 	heta}} - 2 	an 	h

Vedclass · Accepted Answer

$	heta \in \left( 0, \frac{\pi}{2} ight)$

Vedclass · Answer

(A) Given the expression: $\frac{\sin^3 	heta - \cos^3 	heta}{\sin 	heta - \cos 	heta} - \frac{\cos 	heta}{\sqrt{1 + \cot^2 	heta}} - 2 	an 	heta \cot 	heta = -1$Using the identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$,the first term simplifies to $\sin^2 	heta + \sin 	heta \cos 	heta + \cos^2 	heta = 1 + \sin 	heta \cos 	heta$.Since $\sqrt{1 + \cot^2 	heta} = \sqrt{\csc^2 	heta} = |\csc 	heta|$,the second term is $\frac{\cos 	heta}{|\csc 	heta|} = \cos 	heta |\sin 	heta|$.The third term is $2 	an 	heta \cot 	heta = 2(1) = 2$.Substituting these back: $(1 + \sin 	heta \cos 	heta) - \cos 	heta |\sin 	heta| - 2 = -1$,which simplifies to $\sin 	heta \cos 	heta - \cos 	heta |\sin 	heta| = 0$.This implies $\cos 	heta (\sin 	heta - |\sin 	heta|) = 0$.If $\sin 	heta > 0$,then $\sin 	heta - |\sin 	heta| = 0$,which is true for $	heta \in (0, \pi)$.If $\sin 	heta Thus,the condition holds for $	heta \in (0, \pi)$. Comparing with options,option $A$ is the correct subset.

List-$I$	List-$II$
$A$. The period of $\sin^2 x$ is	$I$. $\frac{2\pi}{3}$
$B$. Maximum value of $\frac{\pi}{3}(\sqrt{3}\cos 3x + \sin 3x)$	$II$. $12\pi$
$C$. The period of $\sin \frac{x}{3} + \cos \frac{x}{2}$ is	$III$. $\frac{\pi}{2}$
$D$. Intersection points of $y=\|\sin x\|$ and $y=1$ in $(0, \pi)$	$IV$. $\frac{3\pi}{2}$
	$V$. $\pi$

The equation $\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta - \cos \theta} - \frac{\cos \theta}{\sqrt{1 + \cot^2 \theta}} - 2 \tan \theta \cot \theta = -1$ holds true if:

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