Prove that- sin-6-A-cos-3-B-sin-3-B-cos-6-A-cosA-B

∵sinAcosB+cosAsinB=sin(A+B) So we have, sin[π6+A].cos[π3 B]+sin[π3 B].cos[π6+A] =sin[(π6+A)+(π3 B)] =sin[(π2)+(A B)] =cos(A B) ...

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

Prove that: ...

Question

Prove that:
$sin [\frac{π}{6} + A] . cos [\frac{π}{3} - B] + sin [\frac{π}{3} - B] . cos [\frac{π}{6} + A] = cos (A - B)$

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{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{s i n}^{- 1} (3 x / 5) + {s i n}^{- 1} (4 x / 5) = {s i n}^{- 1} x

{c o s}^{- 1} x + {s i n}^{- 1} (\frac{1}{2} x) = \frac{π}{6}

a \leq {t a n}^{- 1} (\frac{1 - x}{1 + x}) \leq b

0 \leq x \leq 1

(a, b) =

(0, π)

(0, π / 4)

(- π / 4, π / 4)

(π / 4, π / 2)

a \leq {({s i n}^{- 1} x)}^{3} + {({c o s}^{- 1} x)}^{3} \leq b

(\frac{π^{3}}{32}, \frac{7 π^{3}}{8})

\frac{π}{6}

\frac{π}{6}

A = \frac{1}{π} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} {sin}^{- 1} (π x) & {tan}^{- 1} (\frac{π}{x}) {sin}^{- 1} (\frac{π}{x}) & {tan}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦, B = \frac{1}{π} ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} - {cos}^{- 1} (π x) & {tan}^{- 1} (\frac{π}{x}) {sin}^{- 1} (\frac{π}{x}) & - {tan}^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦

A - B

A = \frac{1}{π} ⎡ ⎢ ⎢ ⎣ \begin{matrix} s i n^{- 1} (π x) & t a n^{- 1} (\frac{x}{π}) s i n^{- 1} (\frac{x}{π}) & c o t^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎦, B = ⎡ ⎢ ⎢ ⎣ \begin{matrix} - c o s^{- 1} (π x) & t a n^{- 1} (\frac{x}{π}) s i n^{- 1} (\frac{x}{π}) & t a n^{- 1} (π x) \end{matrix} ⎤ ⎥ ⎥ ⎦

(θ + \frac{π}{3}) = cos (θ - \frac{π}{6}),

tan θ + \sqrt{3} = 0

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