Question

good morning everyone!!!

question for mathematicians ...


the value of tan 1° tan 2° tan 3°.....tan 89° is = ______________


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Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

I'm using basic formula

tan(A + B) = (tan A + tan B) / (1 − tan A tan B)

(1 − tan A tan B) = (tan A + tan B) / tan(A + B)

tan A tan B = 1 - (tan A + tan B) / tan(A + B)

tan 90 = inf

1/tan 90 = 0           [1/inf =0]

tan 1 tan 89 = 1 - (tan 1 + tan 89) / tan 90

                   = 1 - 0 = 1

tan 2 tan 88 = 1 - (tan 2 +tan 88) / tan 90

                    = 1 - 0 = 1

............

.........and so on

tan 44 tan 46 = 1 - (tan 44 tan 46) / tan 90

                      =1 - 0 = 1

Multiplying above all values and tan 45 ( i.e 1 )

we will get

tan 1° tan 2° tan 3°.....tan 89° = 1*1*1*...................1*1 =1

Your answer is 1

Hope it helps :-)