Question

log tan 1 +log tan 2 +log tan 3 ........log tan 89​

Answer 1

Answer:

Step-by-step explanation:

Hi,

logtan1+logtan2+logtan3+……………….+logtan89

=log(tan1×tan2×tan3×……………×tan89)

since loga+logb=log(a×b)

we know that tan89=tan(90-1)=cot1

Tan89=cot1

tan88=cot2

tan87=cot3

.

.

.

.

.

.

We know that tan@×cot@=1

Therefore,

tan1×tan89=1

tan2×tan88=2

.

.

.

.

Therefore the given question will be

log(1×1×1×1×1............×1)=log1=0

                                               OR

we know that log a + log b = log ab.

=>log ( tan 1× tan 2× tan 3 × ............tan 89)

=>log ( tan 1 × tan 2×....tan 44 ×cot 44× cot 43....cot 2 × cot 1). {tan a = cot 90-a}

=> log( 1×1×1×1...×1)

=>log (1)

=> 0.

Read more on Brainly.in - https://brainly.in/question/1088106#readmore

Answer 2

Answer:

log tan 1 + log tan 2 + .....

log tan 89

= log tan (90-1)

= log cot 1

= -log tan 1

similarly every value will get cancelled with its pair

except the middle one which has no pair

therefore ans would be

log tan 45 = log 1 =0

therefore answer is zero

Step-by-step explanation:

thank you

#ckc

plz mark me brainliest

hope it helps