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Given That: (1 + Cos α) (1 + Cos β) (1 + Cos γ) = (1 − Cos α) (1 − Cos α) (1 − Cos β) (1 − Cos γ) Show that One of the Values of Each Member of this Equality is Sin α Sin β Sin γ - Mathematics

Given that:
(1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)

Show that one of the values of each member of this equality is sin α sin β sin γ

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Solution

Given (1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)

Let us assume that

(1 + cos α)(1 + cos β)(1 + cos γ) = (1 -cos α)(1 - cos β)(1 - cos γ) = L

Weknow that `sin^2 theta + cos^2 theta = 1`

Then, we have

L X L = (1 + cos α)(1 +_ cos β)(1 + cos γ) x (1 - cos α)(1 - cos β)(1 - cos γ)

=> :^2 = {(1 - cos α)(1 - cos α)}{(1 + cos β)(1 - cos β)}{(1 + cos γ)(1 - cos γ)}

`=> L^2 = (1 - cos^2 α )(1 - cos^2 β)(1 - cos^2 γ)`

`=> L^2 = sin^2 α sin^2 β sin^2 γ`

`=> L = +- sin α sin β sin γ`

Therefore, we have

`(1 + cos α)(1 + cos β)(1 + cos γ) = (1 - cos α)(1 - cos β)(1 - cos γ) = +- sin α sin β sin γ`

Taking the expression with the positive sign, we have

`(1  + cos α)(1 + cos β)(1 + cos γ) = (1 - cos α)(1 - cos β)(1 - cos γ) = sin α  sin β  sin γ`

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 10 Maths
Chapter 11 Trigonometric Identities
Exercise 11.1 | Q 85 | Page 47
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