If sqrt{2} sec theta + tan theta = 1, then th | vedclass

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(C) Given: $\sqrt{2} \sec 	heta + 	an 	heta = 1$$\Rightarrow \frac{\sqrt{2}}{\cos 	heta} + \frac{\sin 	heta}{\cos 	heta} = 1$$\Rightarrow \sqrt{

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$2n\pi - \frac{\pi}{4}$

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(C) Given: $\sqrt{2} \sec 	heta + 	an 	heta = 1$$\Rightarrow \frac{\sqrt{2}}{\cos 	heta} + \frac{\sin 	heta}{\cos 	heta} = 1$$\Rightarrow \sqrt{2} + \sin 	heta = \cos 	heta$$\Rightarrow \sin 	heta - \cos 	heta = -\sqrt{2}$Dividing both sides by $\sqrt{2}$,we get:$\frac{1}{\sqrt{2}} \sin 	heta - \frac{1}{\sqrt{2}} \cos 	heta = -1$Multiplying by $-1$:$\frac{1}{\sqrt{2}} \cos 	heta - \frac{1}{\sqrt{2}} \sin 	heta = 1$$\Rightarrow \cos 	heta \cos \frac{\pi}{4} - \sin 	heta \sin \frac{\pi}{4} = 1$$\Rightarrow \cos \left( 	heta + \frac{\pi}{4} ight) = 1$Since $\cos \alpha = 1 \Rightarrow \alpha = 2n\pi$,we have:$	heta + \frac{\pi}{4} = 2n\pi$$\Rightarrow 	heta = 2n\pi - \frac{\pi}{4}$.

If $\sqrt{2} \sec \theta + \tan \theta = 1,$ then the general value of $\theta$ is

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