Question

Derive the value of cot45 geometrically

Answer 1

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Q " Derive the Value of Cos 45  Geometrically ?"

Let ΔABC which is right angled at B and ∠BAC = 45°

So , angle ACB is also of 45°

∵∠BAC = ∠ACB = 45°

Then AB = BC

Let AB be x

then BC is also equal to  x

(AB = BC )

Apply pythagoras theorem

AC = $\sqrt{AB^2 + BC^2}$

=> $\sqrt{x^2 + x^2 }$

=> $\sqrt{2x^2}$

AC = $\sqrt{2x}$

We Know that

$Cos \ \theta$ = Base/Hypotenuse

=> B/H

∴ cos∠BAC = AB/AC  ---->  1

Substitute the obtained values in eq (1)

Cos 45° = $\frac{x}{\sqrt{2x} }$

Cos 45° = $\frac{1}{2}$

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Q. Derive the Value of Cot 45 Geometrically ?

=> Cot 45 degree = 1/ tan 45 degree

We Know that ,

tan 45 = 1

Substitute the value

Cot 45 = 1/ 1

Which is equal to 1

Hence , Cot 45 degree have a value of 1

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Answer 2

The value of cot45 geometrically:

1/tan 45 degree

and,

we know that ,

tan 45=1

so,cot 45=1/tan45

cot45=1

Hence proved

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