Question
Evaluate the following:
$\tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ$
$\tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ$
✓
Answer
We have to find $\tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ$
Since $\tan(90^\circ-\theta)=\cot\theta$
So, $\tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ$ $=\tan(90^\circ-42^\circ)\tan(90^\circ-67^\circ)\tan42^\circ\tan67^\circ$
$=\cot42^\circ\cot67^\circ\tan42^\circ\tan67^\circ$
$=(\tan67^\circ\cot67^\circ)(\tan42^\circ\cot42^\circ)$
$=1\times1$
$=1$
So value of $\tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ\text{ is }1$
Since $\tan(90^\circ-\theta)=\cot\theta$
So, $\tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ$ $=\tan(90^\circ-42^\circ)\tan(90^\circ-67^\circ)\tan42^\circ\tan67^\circ$
$=\cot42^\circ\cot67^\circ\tan42^\circ\tan67^\circ$
$=(\tan67^\circ\cot67^\circ)(\tan42^\circ\cot42^\circ)$
$=1\times1$
$=1$
So value of $\tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ\text{ is }1$
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