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Trigonometric Ratios

Given that:1+...

Question

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

sin α sin β sin γ .

∣ ∣ ∣ ∣ \begin{matrix} 1 + c o s α & 1 + s i n α & 1 1 + c o s β & 1 + s i n β & 1 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣

\neq

c o t^{- 1} [(c o s α)^{\frac{1}{2}}] - t a n^{- 1} [(c o s α)^{\frac{1}{2}}] = x

s i n x =

a \times c o s β = c o s α

b \times s i n β = c o s α

(a + b) c o s β

(\frac{1 + sin α}{1 + cos α}) (\frac{1 + sec α}{1 + c o s e c α}) = tan α

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