Question

Prove that tan 1° tan 2° tan 3° ... tan 89° = 1

Answer 1

Answer:

the answer for the given problem is 1

Answer 2

Answer:

We have

tan 1° tan 2° tan 3°... tan 89° [LHS]

= tan 1°tan 2°... tan 44° tan 45° tan 46°... tan 88° tan 89°

= (tan 1°tan 89°)(tan 2° tan 88°)... (tan 44° tan 46°) tan 45°

= {(tan 1° tan (90° -1°)} . {tan 2° tan (90° -2°)} ...{tan 44° tan (90°-44°)} tan 45°*

= (tan 1°cot 1°)(tan 2°cot 2°)... (tan 44°cot 44°).1

[°.° tan (90°-8)= cot 0 and tan 45° = 1]

= 1×1×...×1×1

= 1.

LHS = RHS

$\:$

Hence, proved!