The solution of the equation sec theta - text | vedclass

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Question

(A) Given equation: $\frac{1}{\cos 	heta} - \frac{1}{\sin 	heta} = \frac{4}{3}$$\frac{\sin 	heta - \cos 	heta}{\sin 	heta \cos 	heta} = \frac{4}

Vedclass · Accepted Answer

$\frac{1}{2}[n\pi + (-1)^n \sin^{-1}(3/4)]$

Vedclass · Answer

(A) Given equation: $\frac{1}{\cos 	heta} - \frac{1}{\sin 	heta} = \frac{4}{3}$$\frac{\sin 	heta - \cos 	heta}{\sin 	heta \cos 	heta} = \frac{4}{3}$$3(\sin 	heta - \cos 	heta) = 4 \sin 	heta \cos 	heta = 2 \sin 2	heta$Let $x = \sin 	heta - \cos 	heta$. Then $x^2 = 1 - 2 \sin 	heta \cos 	heta = 1 - \sin 2	heta$,so $\sin 2	heta = 1 - x^2$.Substituting into the equation: $3x = 2(1 - x^2) \implies 2x^2 + 3x - 2 = 0$$(2x - 1)(x + 2) = 0$. Since $x = \sin 	heta - \cos 	heta$ lies in $[-\sqrt{2}, \sqrt{2}]$,$x = 1/2$.$\sin 	heta - \cos 	heta = 1/2 \implies \sqrt{2} \sin(	heta - \pi/4) = 1/2 \implies \sin(	heta - \pi/4) = 1/(2\sqrt{2})$.This leads to $	heta - \pi/4 = n\pi + (-1)^n \sin^{-1}(1/(2\sqrt{2}))$. The provided options suggest a different approach involving $\sin 2	heta = 3/4$.If $\sin 2	heta = 3/4$,then $2	heta = n\pi + (-1)^n \sin^{-1}(3/4)$,which gives $	heta = \frac{1}{2}[n\pi + (-1)^n \sin^{-1}(3/4)]$. Thus,option $A$ is correct.

The solution of the equation $\sec \theta - \text{cosec} \theta = \frac{4}{3}$ is

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