If tan theta - sqrt{2} sec theta = sqrt{3},th | vedclass

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(D) Given: $	an 	heta - \sqrt{2} \sec 	heta = \sqrt{3}$ Multiply by $\cos 	heta$: $\sin 	heta - \sqrt{2} = \sqrt{3} \cos 	heta$ Rearrange: $\sin

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$n\pi + (-1)^n \frac{\pi}{4} + \frac{\pi}{3}$

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(D) Given: $	an 	heta - \sqrt{2} \sec 	heta = \sqrt{3}$ Multiply by $\cos 	heta$: $\sin 	heta - \sqrt{2} = \sqrt{3} \cos 	heta$ Rearrange: $\sin 	heta - \sqrt{3} \cos 	heta = \sqrt{2}$ Divide by $2$: $\frac{1}{2} \sin 	heta - \frac{\sqrt{3}}{2} \cos 	heta = \frac{\sqrt{2}}{2}$ Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $\sin(	heta - \frac{\pi}{3}) = \sin(\frac{\pi}{4})$ The general solution for $\sin x = \sin \alpha$ is $x = n\pi + (-1)^n \alpha$. Therefore,$	heta - \frac{\pi}{3} = n\pi + (-1)^n \frac{\pi}{4}$ $	heta = n\pi + (-1)^n \frac{\pi}{4} + \frac{\pi}{3}$.

$List-I$	$List-II$
$(I)$ $\{x \in[-\frac{2 \pi}{3}, \frac{2 \pi}{3}]: \cos x+\sin x=1\}$	$(P)$ has two elements
$(II)$ $\{x \in[-\frac{5 \pi}{18}, \frac{5 \pi}{18}]: \sqrt{3} \tan 3 x=1\}$	$(Q)$ has three elements
$(III)$ $\{x \in[-\frac{6 \pi}{5}, \frac{6 \pi}{5}]: 2 \cos (2 x)=\sqrt{3}\}$	$(R)$ has four elements
$(IV)$ $\{x \in[-\frac{7 \pi}{4}, \frac{7 \pi}{4}]: \sin x-\cos x=1\}$	$(S)$ has five elements
	$(T)$ has six elements

If $\tan \theta - \sqrt{2} \sec \theta = \sqrt{3}$,then the general value of $\theta$ is

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