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Question: If \(0 <\alpha, \beta, \gamma < \dfrac{\pi }{2}\) such that \[\alpha + \beta + \gamma = \dfrac{\pi }...

If 0<α,β,γ<π20 <\alpha, \beta, \gamma < \dfrac{\pi }{2} such that α+β+γ=π2\alpha + \beta + \gamma = \dfrac{\pi }{2} and cot⁡α,cot⁡β,cot⁡γ\cot \alpha ,\cot \beta ,\cot \gamma are in A.P., then the value of cot⁡αcot⁡γ\cot \alpha \cot \gamma is:

Explanation

Solution

Arithmetic progression (AP) is a sequence whose terms increase or decrease by a fixed number. The fixed number is called the common difference. If ‘a’ is the first term and ‘d’ is a common difference, then AP can be written as a, a+d, a+2d,….a+(n-1)d,…..
For example, if x, y, z are in AP then it implies that x + z = 2y
Given: α+β+γ=π2\alpha + \beta + \gamma = \dfrac{\pi }{2} and cot⁡α,cot⁡β,cot⁡γ\cot \alpha ,\cot \beta ,\cot \gamma are in AP
To find: the value of cot⁡αcot⁡γ\cot \alpha \cot \gamma

Complete step-by-step solution:
Step 1: As we know that if three numbers, let say x, y and z, are in AP then it implies that x+z=2yx + z = 2y
Now according to the question cot⁡α,cot⁡β,cot⁡γ\cot \alpha ,\cot \beta ,\cot \gamma are in AP, therefore we can write
cot⁡α+cot⁡γ=2cot⁡β\cot \alpha + \cot \gamma = 2\cot \beta
Also, it is given that α+β+γ=π2\alpha + \beta + \gamma = \dfrac{\pi }{2} , we can also write it as
β=π2−(α+β)\beta = \dfrac{\pi }{2} - (\alpha + \beta )
Now substituting the value of β\beta in equation cot⁡α+cot⁡γ=2cot⁡β\cot \alpha + \cot \gamma = 2\cot \beta , we get
cot⁡α+cot⁡γ=2cot⁡β\cot \alpha + \cot \gamma = 2\cot \beta
cot⁡α+cot⁡γ=2cot⁡(π2−(α+γ))\cot \alpha + \cot \gamma = 2\cot (\dfrac{\pi }{2} - (\alpha + \gamma ))
Step 2: Now from trigonometric properties we know that,
cot⁡(π2−x)=tan⁡x\cot (\dfrac{\pi }{2} - x) = \tan x
Therefore, we have
cot⁡α+cot⁡γ=2cot⁡(π2−(α+γ))\cot \alpha + \cot \gamma = 2\cot (\dfrac{\pi }{2} - (\alpha + \gamma ))
cot⁡α+cot⁡γ=2tan⁡(α+γ)\cot \alpha + \cot \gamma = 2\tan (\alpha + \gamma )
1tan⁡α+1tan⁡γ=2tan⁡(α+γ)\dfrac{1}{{\tan \alpha }} + \dfrac{1}{{\tan \gamma }} = 2\tan (\alpha + \gamma )
Step 3: As we know that,
tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}
Using the above trigonometric formula, we have
1tan⁡α+1tan⁡γ=2tan⁡(α+γ)\dfrac{1}{{\tan \alpha }} + \dfrac{1}{{\tan \gamma }} = 2\tan (\alpha + \gamma )
1tan⁡α+1tan⁡γ=2tan⁡α+tan⁡γ1−tan⁡αtan⁡γ\dfrac{1}{{\tan \alpha }} + \dfrac{1}{{\tan \gamma }} = 2\dfrac{{\tan \alpha + \tan \gamma }}{{1 - \tan \alpha \tan \gamma }}
On further simplification we get
tan⁡α+tan⁡γtan⁡αtan⁡γ=2tan⁡α+tan⁡γ1−tan⁡αtan⁡γ\dfrac{{\tan \alpha + \tan \gamma }}{{\tan \alpha \tan \gamma }} = 2\dfrac{{\tan \alpha + \tan \gamma }}{{1 - \tan \alpha \tan \gamma }}
Canceling the term tan⁡α+tan⁡γ\tan \alpha + \tan \gamma from both sides, we get
1tan⁡αtan⁡γ=211−tan⁡αtan⁡γ\dfrac{1}{{\tan \alpha \tan \gamma }} = 2\dfrac{1}{{1 - \tan \alpha \tan \gamma }}
1−tan⁡αtan⁡γ=2tan⁡αtan⁡γ1 - \tan \alpha \tan \gamma = 2\tan \alpha \tan \gamma
Taking the term tan⁡αtan⁡γ\tan \alpha \tan \gamma on the right side, we get
1=2tan⁡αtan⁡γ+tan⁡αtan⁡γ1 = 2\tan \alpha \tan \gamma + \tan \alpha \tan \gamma
1=3tan⁡αtan⁡γ1 = 3\tan \alpha \tan \gamma
Now from trigonometric properties, we know that cot⁡x=1tan⁡x\cot x = \dfrac{1}{{\tan x}} , using this trigonometric property we can write
1=3tan⁡αtan⁡γ1 = 3\tan \alpha \tan \gamma
1=31cot⁡αcot⁡γ1 = 3\dfrac{1}{{\cot \alpha \cot \gamma }}
cot⁡αcot⁡γ=3\cot \alpha \cot \gamma = 3 (which is the required answer)
Hence, the value cot⁡αcot⁡γ\cot \alpha \cot \gamma is equal to 3.

Note: Always use trigonometric properties to convert the given equation into one identity (tan or cot or sin or cos etc) and also it reduces the equation in the simplest form.
If three numbers a, b and c are in AP then it implies that a+c=2ba + c = 2b.