Question
Evaluate the following:
$\frac{\cot40^\circ}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\cos35^\circ}{\sin55^\circ}\Big)$
$\frac{\cot40^\circ}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\cos35^\circ}{\sin55^\circ}\Big)$
✓
Answer
We have to find: $\frac{\cot40^\circ}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\cos35^\circ}{\sin55^\circ}\Big)$$$
Since $\cot(90^\circ-\theta)=\tan\theta$ and $\cos(90^\circ-\theta)=\sin\theta$
So, $\frac{\cot40^\circ}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\cos35^\circ}{\sin55^\circ}\Big)=\frac{(\cot90^\circ-50^\circ)}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\cos(90^\circ-55^\circ)}{\sin55^\circ}\Big)$
$=\frac{\tan50^\circ}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\sin55^\circ}{\sin55^\circ}\Big)$
$=1-\frac{1}{2}$
$=\frac{1}{2}$
So value of $\frac{\cot40^\circ}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\cos35^\circ}{\sin55^\circ}\Big)\text{ is }\frac{1}{2}$
Since $\cot(90^\circ-\theta)=\tan\theta$ and $\cos(90^\circ-\theta)=\sin\theta$
So, $\frac{\cot40^\circ}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\cos35^\circ}{\sin55^\circ}\Big)=\frac{(\cot90^\circ-50^\circ)}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\cos(90^\circ-55^\circ)}{\sin55^\circ}\Big)$
$=\frac{\tan50^\circ}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\sin55^\circ}{\sin55^\circ}\Big)$
$=1-\frac{1}{2}$
$=\frac{1}{2}$
So value of $\frac{\cot40^\circ}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\cos35^\circ}{\sin55^\circ}\Big)\text{ is }\frac{1}{2}$
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