Given that-1-cos-1-cos-1-cos-1-cos-1-cos-1-cos- Show that one of the values of each member of this equality is sin-sin-sin-

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Trigonometric Identities

Given that:1+...

Question

$Given that : (1 + c o s α) (1 + c o s β) (1 + c o s γ) = (1 - c o s α) (1 - c o s β) (1 - c o s γ)$

Show that one of the values of each member of this equality is $s i n α s i n β s i n γ$

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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

\cos θ = \frac{\cos α + \cos β}{1 + \cos α \cos β}

\tan \frac{θ}{2} = \pm \tan \frac{α}{2} \tan \frac{β}{2}

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