Evaluate:\( \dfrac{sin^263° + sin^227°}{cos^217° + cos^273°}\)

Asked by Abhisek | 1 year ago |  98

1 Answer

Solution :-

\( \dfrac{(sin^263° + sin^227°)}{(cos^217° + cos^273°)}\)

To simplify this, convert some of the sin functions into cos 

functions and cos function into sin function and it becomes,

\( \dfrac{[sin^2(90°-27°) + sin^227°]}{[cos^2(90°-73°) + cos^273°]}\)

\( \dfrac{(cos^227° + sin^227°)}{(sin^227° + cos^273°)}\)

\(\dfrac{1}{1}\) =1  (since sin2A + cos2A = 1)

Therefore, \( \dfrac{(sin^263° + sin^227°)}{(cos^217° + cos^273°)}=1\)

Answered by Pragya Singh | 1 year ago

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