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

Asked by Abhisek | 1 year ago |  98

##### 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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