frac{sin^3 theta - cos^3 theta}{sin theta - c | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given 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 $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$,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 $\frac{\cos 	heta}{|\csc 	heta|} = \cos 	heta |\sin 	heta|$.Also,$2 	an 	heta \cot 	heta = 2(1) = 2$.Substituting these: $(1 + \sin 	heta \cos 	heta) - \cos 	heta |\sin 	heta| - 2 = -1$.$\sin 	heta \cos 	heta - \cos 	heta |\sin 	heta| = 0$.$\cos 	heta (\sin 	heta - |\sin 	heta|) = 0$.This holds if $\cos 	heta = 0$ or $\sin 	heta = |\sin 	heta|$.$\sin 	heta = |\sin 	heta|$ implies $\sin 	heta \ge 0$,which occurs in the first and second quadrants,i.e.,$	heta \in (0, \pi)$.However,the denominator $\sin 	heta - \cos 	heta 
eq 0$ implies $	heta 
eq \frac{\pi}{4}, \frac{5\pi}{4}$.Checking the options,$	heta \in (0, \pi/2)$ satisfies $\sin 	heta > 0$,making the expression $1 + \sin 	heta \cos 	heta - \cos 	heta \sin 	heta - 2 = -1$.

$\frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta - \cos \theta} - \frac{\cos \theta}{\sqrt{1 + \cot^2 \theta}} - 2 \tan \theta \cot \theta = -1$ if

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