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The value of logtan1o+logtan2o+.....+logtan89olog tan 1^o + logtan 2^o + ..... + logtan89^o logtan1o+logtan2o+.....+logtan89o is :​
Question

The value of logtan1o+logtan2o+.....+logtan89olog tan 1^o + logtan 2^o + ..... + logtan89^o  is :​

A.

-1

B.

0

C.

1

D.

π2\frac{​π}{2​}

Correct option is B

Given:

logtan1o+logtan2o+.....+logtan89olog tan 1^o + logtan 2^o + ..... + logtan89^o

Formula Used:

logm+logn=log(m×n)log m + log n = log (m \times n)

tanθ=cot(90θ)tan \theta = cot (90-\theta)

tanθ×cotθ=1tan \theta \times cot \theta = 1

loga1=0log a^1 = 0​​

Solution:

tan89=cot(9089)=cot1\tan 89^\circ = \cot (90^\circ - 89^\circ) = \cot 1^\circ

tan88=cot2,\tan 88^\circ = \cot 2^\circ,

tan87=cot3\tan 87^\circ = \cot 3^\circ ...

By logm+logn=log(m×n)log m + log n = log (m \times n)

logtan1+logtan89\log \tan 1^\circ + \log \tan 89^\circ

=log(tan1×tan89)= \log (\tan 1^\circ \times \tan 89^\circ)​​

=log(tan1×cot1)= \log (\tan 1^\circ \times \cot 1^\circ)

​​=log1= \log 1

= 0 

Thus, the correct answer is (b).

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