Question

Evaluate tan15°tan35°tan30°tan65°tan85° .

Answer 1

Here make use of the identities,

$\tan (90^o-\theta)=\cot \theta =\frac{1}{\tan \theta} \\ \tan (90^o-\theta)\tan \theta=1$

Therefore,

$\tan (15^o)\tan (85^o)=1\\ \tan (35^o)\tan (65^o)=1$

The given expression is

$\tan (15^o)\tan (85^o)\tan (35^o)\tan (65^o)\tan (30^o)=1*1*\tan (30^o)=\tan (30^o)=\frac{1}{\sqrt{3}}$