Question
Without using trigomentric tables, prove that:
$\tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ=1$
$\tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ=1$
✓
Answer
$\text{L.H.S}=\tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ$
$=\cot\big(90^\circ+48^\circ\big)\cot\big(90^\circ-23^\circ\big)-\tan42^\circ\tan67^\circ$
$=\cot42^\circ\cot67^\circ\tan42^\circ\tan67^\circ$
$=\frac{1}{\tan42^\circ}\times\frac{1}{\tan67^\circ}\times\tan42^\circ\times\tan67^\circ$
$=1$
$=\text{R.H.S.}$
$=\cot\big(90^\circ+48^\circ\big)\cot\big(90^\circ-23^\circ\big)-\tan42^\circ\tan67^\circ$
$=\cot42^\circ\cot67^\circ\tan42^\circ\tan67^\circ$
$=\frac{1}{\tan42^\circ}\times\frac{1}{\tan67^\circ}\times\tan42^\circ\times\tan67^\circ$
$=1$
$=\text{R.H.S.}$
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