Question

find the value of cot1,cot2,cot3.............................cot89​

Answer 1

Answer:1

Step-by-step explanation:

cot1.cot2.cot3 ……………..cot87.cot88.cot89

we know,

cot1 =cot(90-89) = tan89

cot1.cot89 =1

in the same way ,

cot2.cot88 =1

core.cot87 =1

..…………

{cot1.cot89}{cot2.cot88}{cot3.cot87}.……

=1 × 1 × 1 × 1 …………

=1

Answer 2

Answer:

The value of the series is 1.

Step-by-step explanation:

• cot1,cot2,cot3.............................cot89​

The given equation can be written as

$\left(\cot 1^{\circ} \cot 89^{\circ}\right)\left(\cot 2^{\circ} \cot 88^{\circ}\right) \ldots . .\left(\cot 44^{\circ} \cot 46^{\circ}\right) \cot 45^{\circ}$

$=\left(\cot 1^{\circ} \cot \left(90^{\circ}-1^{\circ}\right)\right)\left(\cot 2^{\circ} \cot \left(90^{\circ}-2^{\circ}\right)\right) \ldots . .\left(\cot 44^{\circ} \cot \left(90^{\circ}-44^{\circ}\right)\right) \cot 45^{\circ}$

The given equation can be written as

$\left(\cot 1^{\circ} \cot 89^{\circ}\right)\left(\cot 2^{\circ} \cot 88^{\circ}\right) \ldots .\left(\cot 44^{\circ} \cot 46^{\circ}\right) \cot 45^{\circ}$

$=\left(\cot 1^{\circ} \cot \left(90^{\circ}-1^{\circ}\right)\right)\left(\cot 2^{\circ} \cot \left(90^{\circ}-2^{\circ}\right)\right) \ldots .\left(\cot 44^{\circ} \cot \left(90^{\circ}-44^{\circ}\right)\right) \cot 45^{\circ} =\cot 1^{\circ} \tan 1^{\circ} \cdot \cot 2^{\circ} \tan 2^{\circ} \ldots . \cot 44^{\circ} \tan 44^{\circ} \times \cot 45^{\circ} \quad\left(\right. Since \left.\left.\cot \left(90^{\circ}-x\right)=\tan x\right)\right)$$=1 \\=\cot 1^{\circ} \tan 1^{\circ} \cdot \cot 2^{\circ} \tan 2^{\circ} \ldots . \cot 44^{\circ} \tan 44^{\circ} \times \cot 45^{\circ} \quad\left(\right. Since \left.\left.\cot \left(90^{\circ}-x\right)=\tan x\right)\right) =1 \times 1 \ldots .1 \times 1 \quad\left(\right. Since \left.\cot x \times \tan x=1, \cot 45^{\circ}\right) =1$