If |sin alpha - cos alpha| = frac{3}{4},then | vedclass

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(C) Given $|\sin \alpha - \cos \alpha| = \frac{3}{4}$.Squaring both sides,we get $(\sin \alpha - \cos \alpha)^2 = (\frac{3}{4})^2$.$\sin^2 \alpha + \c

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$\frac{3}{\sqrt{23}}$

Vedclass · Answer

(C) Given $|\sin \alpha - \cos \alpha| = \frac{3}{4}$.Squaring both sides,we get $(\sin \alpha - \cos \alpha)^2 = (\frac{3}{4})^2$.$\sin^2 \alpha + \cos^2 \alpha - 2 \sin \alpha \cos \alpha = \frac{9}{16}$.$1 - \sin 2\alpha = \frac{9}{16}$,which implies $\sin 2\alpha = 1 - \frac{9}{16} = \frac{7}{16}$.Since $\sin 2\alpha = \frac{7}{16}$,we have the perpendicular $P = 7$ and hypotenuse $H = 16$.The base $B = \sqrt{H^2 - P^2} = \sqrt{16^2 - 7^2} = \sqrt{256 - 49} = \sqrt{207} = 3\sqrt{23}$.Thus,$\cos 2\alpha = \frac{B}{H} = \frac{3\sqrt{23}}{16}$.Now,$|\sec 2\alpha - 	an 2\alpha| = |\frac{1}{\cos 2\alpha} - \frac{\sin 2\alpha}{\cos 2\alpha}| = |\frac{1 - \sin 2\alpha}{\cos 2\alpha}|$.Substituting the values,$|\frac{1 - 7/16}{3\sqrt{23}/16}| = |\frac{9/16}{3\sqrt{23}/16}| = \frac{9}{3\sqrt{23}} = \frac{3}{\sqrt{23}}$.Therefore,option $C$ is correct.

If $|\sin \alpha - \cos \alpha| = \frac{3}{4}$,then $|\sec 2\alpha - \tan 2\alpha| = $

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