hloo guysplzzz explain this properly in radians. Created by weirdgirl66. Math

hloo guysplzzz explain this properly in radians...

vrrunda Cos¢=1/√2. sin¢=-1/√2

here cos¢=(+)ve in first and fourth quadrant.

sin¢(-)ve in third and fourth quadrant .

so we can write that ¢ is in 4 th quadrant.

in fourth quadrant cos (360-¢)=cos¢=cosπ/4

so. cos(2π-¢)=cosπ/4

2π-¢=π/4

¢=2π-π/4=7π/4

vrrunda plz make me as brainliest answer

rakeshmohata Hope u like my process
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Given
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$= > \it \sin( \theta ) = - \frac{1}{ \sqrt{2} } \\ \\ \it = > \cos( \theta ) = \frac{1}{ \sqrt{2} } \\ \\ \bf \: \: \: \: \: \: \: \: \: \underline{...now...} \\ \\ dividing \: \: \sin( \theta ) \: \: by \: \: \cos( \theta ) \\ \\ = > \frac{ \sin( \theta ) }{ \cos( \theta ) } = \blue{ \frac{ - \frac{1}{ \sqrt{2} } }{ \frac{1}{ \sqrt{2} } }} = \blue{ - 1} \\ \\ = > \tan( \theta ) = \blue{ - 1} = \blue{\tan(n\pi - \frac{\pi}{4} ) }\\ \\ = > \boxed{ \bf \: \theta = \underline{ \green{n\pi - \frac{\pi}{4}} }}$

Where n =0,1,2,3,...
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