Cot -4 - - - cot -4 - - -

The correct option is C 1 (π4 + θ) . (π4 θ) = (π4 +θ) [π2 (π4 +θ)] = (π4 +θ) tan (π4 +θ) #160; #160; #160; [ (π2 x )=tan x] ...

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Sin(A+B)Sin(A-B)

cot π/4 + θ ....

Question

$c o t (\frac{π}{4} + θ) . c o t (\frac{π}{4} - θ) =$

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c o s e c \frac{π}{32} + c o s e c \frac{π}{16} + c o s e c \frac{π}{8} + c o s e c \frac{π}{4} + c o s e c \frac{π}{2} = c o t \frac{π}{k}

The value of $c o t (\frac{π}{4} - 2 c o t^{- 1} 3)$ is equal to

θ

θ = 0

θ = π / 2

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + s i n^{2} θ & c o s^{2} θ & 4 s i n 4 θ s i n^{2} θ & 1 + c o s^{2} θ & 4 s i n 4 θ s i n^{2} θ & c o s^{2} θ & 1 + 4 s i n 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

\frac{2 s i n^{2} (\frac{π}{4} - α)}{c o s 2 α} = c o t (\frac{π}{4} + α) .

θ

θ = 0

\frac{π}{2}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + {sin}^{2} θ & {cos}^{2} θ & 4 sin 4 θ {sin}^{2} θ & 1 + {cos}^{2} θ & 4 sin 4 θ {sin}^{2} θ & {cos}^{2} θ & 1 + 4 sin 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

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