If tan theta = frac{{sin alpha - cos alpha }} | vedclass

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Question

(a) We have $	an 	heta = \frac{{\sin \alpha - \cos \alpha }}{{\sin \alpha + \cos \alpha }}$

$ \Rightarrow 	an 	heta = \frac{{\sin \left( {\alph

Vedclass · Accepted Answer

$\sqrt 2 \cos 	heta ,\,\,\sqrt 2 \sin 	heta $

Vedclass · Answer

(a) We have $	an 	heta = \frac{{\sin \alpha - \cos \alpha }}{{\sin \alpha + \cos \alpha }}$

$ \Rightarrow 	an 	heta = \frac{{\sin \left( {\alpha - \frac{\pi }{4}} ight)}}{{\cos \left( {\alpha - \frac{\pi }{4}} ight)}} $

$\Rightarrow 	an 	heta = 	an \left( {\alpha - \frac{\pi }{4}} ight)$

$ \Rightarrow 	heta = \alpha - \frac{\pi }{4}$

$\Rightarrow \alpha = 	heta + \frac{\pi }{4}$

Hence, $\sin \alpha + \cos \alpha = \sin \left( {	heta + \frac{\pi }{4}} ight) + \cos \left( {	heta + \frac{\pi }{4}} ight)$

$ = \sqrt 2 \cos 	heta $

and $\sin \alpha - \cos \alpha = \sin \left( {	heta + \frac{\pi }{4}} ight) - \cos \left( {	heta + \frac{\pi }{4}} ight)$

$ = \frac{1}{{\sqrt 2 }}\sin 	heta + \frac{1}{{\sqrt 2 }}\cos 	heta - \frac{1}{{\sqrt 2 }}\cos 	heta + \frac{1}{{\sqrt 2 }}\sin 	heta $

$ = \frac{2}{{\sqrt 2 }}\sin 	heta = \sqrt 2 \sin 	heta $.

If $\tan \theta = \frac{{\sin \alpha - \cos \alpha }}{{\sin \alpha + \cos \alpha }},$ then $\sin \alpha + \cos \alpha $ and $\sin \alpha - \cos \alpha $ must be equal to

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