cot10.cot20.cot60.cot70.cot80 is equal to ?
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Answered by
23
Heya !
☆ here is your answer ☆
cot70.cot20× cot80. cot10× cot60
(tan90- 70).cot20 × tan (90- 80)cot10× cot 60
tan20× cot20 ×tan10× cot10×cot 60
tan20× 1/tan20 × tan 10× 1/tan10× cot 60
1 x 1 ×1/root 3
1 ×1 / root 3 = 1 /root 3
☆ here is your answer ☆
cot70.cot20× cot80. cot10× cot60
(tan90- 70).cot20 × tan (90- 80)cot10× cot 60
tan20× cot20 ×tan10× cot10×cot 60
tan20× 1/tan20 × tan 10× 1/tan10× cot 60
1 x 1 ×1/root 3
1 ×1 / root 3 = 1 /root 3
arjunkumar45:
thanks
Answered by
6
HEYA MATE, HERE IS UR ANSWER

Rearranging so that calculation becomes easier
= Cot 70 ×cot 20×cot 10× cot 80 ×cot 60
We know the identity cot (90-A)= cot A
we will use this identity here
Cot 70 ×cot (90-70)×cot 80 ×cot (90-80)×cot 60
cot 70×Tan 70×cot 80×Tan 80 ×cot 60
Now we know that Tan A=
Cot 70×
×Cot 80 ×
×cot 60
All the like terms will cancel and cot 60 will remain
We know that cot 60=
So ur answer is
Rearranging so that calculation becomes easier
= Cot 70 ×cot 20×cot 10× cot 80 ×cot 60
We know the identity cot (90-A)= cot A
we will use this identity here
Cot 70 ×cot (90-70)×cot 80 ×cot (90-80)×cot 60
cot 70×Tan 70×cot 80×Tan 80 ×cot 60
Now we know that Tan A=
Cot 70×
All the like terms will cancel and cot 60 will remain
We know that cot 60=
So ur answer is
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