Question
Evaluate the following:
$\frac{\tan35^\circ}{\cot55^\circ}+\frac{\cot78^\circ}{\tan12^\circ}-1$
$\frac{\tan35^\circ}{\cot55^\circ}+\frac{\cot78^\circ}{\tan12^\circ}-1$
✓
Answer
We have to find: $\frac{\tan35^\circ}{\cot55^\circ}+\frac{\cot78^\circ}{\tan12^\circ}-1$$$ Since $\tan(90^\circ-\theta)=\cot\theta$ and $\cot(90^\circ-\theta)=\tan\theta$ $\frac{\tan(90-55)}{\cot55}+\frac{\cot(90-12)}{\tan12}-1$$\frac{\cot55^\circ}{\cot55^\circ}+\frac{\tan12^\circ}{\tan12^\circ}-1$
$1+1-1=1$
$1+1-1=1$
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