Is arctan(x)=(arcsin(x))/(arccos(x)) ?

The answer to whether arctan(x)=(arcsin(x))/(arccos(x)) is False

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prove arctan(x)=(arcsin(x))/(arccos(x))

prove

arctan (x) = \frac{arcsin ( x )}{arccos ( x )}

False

Solution steps

arctan (x) = \frac{arcsin ( x )}{arccos ( x )}

Show that the two sides are not equal

For

arctan (x) = \frac{arcsin ( x )}{arccos ( x )}

plug in

x = \frac{1}{2}

arctan (\frac{1}{2}) = 0.46364 \dots (Degrees : 26.5 7^{\circ})

arctan (\frac{1}{2})

Refine to a decimal form

= 0.46364 \dots

\frac{arcsin ( \frac{1}{2} )}{arccos ( \frac{1}{2} )} = \frac{1}{2} (Decimal : Degrees : 0.5 28.6 5^{\circ})

\frac{arcsin ( \frac{1}{2} )}{arccos ( \frac{1}{2} )}

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{6}

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{π}{3}

arccos (\frac{1}{2})

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{3}

= \frac{\frac{π}{6}}{\frac{π}{3}}

Simplify

\frac{\frac{π}{6}}{\frac{π}{3}} : \frac{1}{2}

\frac{\frac{π}{6}}{\frac{π}{3}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{π 3}{6 π}

Cancel the common factor:

π

= \frac{3}{6}

Cancel the common factor:

3

= \frac{1}{2}

= \frac{1}{2}

The two sides are not equal

\Rightarrow False

p ro v e \frac{csc ( θ ) + 1}{1 + sin ( θ )} = csc (θ)

p ro v e sec (A) csc (A) = csc (A) tan (A)

p ro v e sin (x) \cdot tan (x) + cos (- x) = sec (x)

p ro v e cos (2 α) = cos^{2} (α) - sin^{2} (α)

p ro v e \frac{tan ( x )}{1 + tan ^{2} ( x )} = cos (x) sin (x)

Is arctan(x)=(arcsin(x))/(arccos(x)) ?
The answer to whether arctan(x)=(arcsin(x))/(arccos(x)) is False