Question

Solve:

$\tan 1^o.\tan 2^o.\tan 3^o......\tan {89}^o=$

• A. $2$
• B. $3$
• C. $1$
• D. $4$

Answer 1

The correct option is C $1$

$tan{1}^{\circ }.tan{2}^{\circ }.tan{3}^{\circ }...tan{89}^{\circ }$

$=tan{1}^{\circ }.tan{2}^{\circ }.tan{3}^{\circ }...cot{1}^{\circ }cot{2}^{\circ }cot{3}^{\circ }...$

$\left(tan\right(90-0)=cot\theta )$

$=\mathrm{1.....}(tan\theta \times cot\theta )=1$