If theta and phi are angles in the 1^{st} q | vedclass

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(d) Given, $	an 	heta = \frac{1}{7},\sin \phi = \frac{1}{{\sqrt {10} }}$

$\sin 	heta = \frac{1}{{\sqrt {50} }},\,\,\cos 	heta = \frac{7}{{

Vedclass · Accepted Answer

$	heta + 2\phi = 45^\circ $

Vedclass · Answer

(d) Given, $	an 	heta = \frac{1}{7},\sin \phi = \frac{1}{{\sqrt {10} }}$

$\sin 	heta = \frac{1}{{\sqrt {50} }},\,\,\cos 	heta = \frac{7}{{\sqrt {50} }},\,\,\cos \phi = \frac{3}{{\sqrt {10} }}$

$	herefore \,\,\cos 2\phi  = 2{\cos ^2}\phi - 1 = 2.\frac{9}{{10}} - 1 = \frac{8}{{10}}$

$\sin 2\phi = 2\sin \phi \cos \phi = 2 	imes .\frac{1}{{\sqrt {10} }} 	imes \frac{3}{{\sqrt {10} }} = \frac{6}{{10}}$

$	herefore \cos (	heta + 2\phi ) = \cos 	heta \cos 2\phi - \sin 	heta \sin 2\phi $

$ = \frac{7}{{\sqrt {50} }} 	imes \frac{8}{{10}} - \frac{1}{{\sqrt {50} }}.\frac{6}{{10}}$

$ = \frac{{56 - 6}}{{10\sqrt {50} }} = \frac{{50}}{{10\sqrt {50} }} = \frac{{5\sqrt 2 }}{{10}} = \frac{1}{{\sqrt 2 }}$

$	herefore 	heta + 2\phi = {45^o}$.

If $\theta $ and $\phi $ are angles in the $1^{st}$ quadrant such that $\tan \theta = 1/7$ and $\sin \phi = 1/\sqrt {10} $.Then

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