Question

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

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

Answer: (C)

$1$

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

$=\tan 1^{\circ }\tan 2^{\circ }\tan 3^{\circ }...\tan {87}^{\circ }\tan {88}^{\circ }\tan {89}^{\circ }$

$=\tan 1^{\circ }\tan 2^{\circ }\tan 3^{\circ }...\tan ({90}^{\circ }-3^{\circ })\tan ({90}^{\circ }-2^{\circ })\tan ({90}^{\circ }-1^{\circ })$

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

$=\left(\tan 1^{\circ }\cot 1^{\circ }\right)\left(\tan 2^{\circ }\cot 2^{\circ }\right)\left(\tan 2^{\circ }\cot 2^{\circ }\right)...\left(\tan {45}^{\circ }\cot {45}^{\circ }\right)$ since $\tan \theta \cot \theta =1$

$=1\times 1\times 1\times ...\times 1$

$=1$