What is the value of [ln(0+1)]cosh(0)+tan(0) ?

The value of [ln(0+1)]cosh(0)+tan(0) is 0

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[ln(0+1)]cosh(0)+tan(0)

[ln (0 + 1)] cosh (0) + tan (0)

0

Solution steps

[ln (0 + 1)] cosh (0) + tan (0)

[ln (0 + 1)] cosh (0) + tan (0) = tan (0)

(ln (0 + 1)) cosh (0) + tan (0)

Remove parentheses:

(a) = a

= ln (0 + 1) cosh (0) + tan (0)

ln (0 + 1) cosh (0) = 0

ln (0 + 1) cosh (0)

Add the numbers:

0 + 1 = 1

= ln (1) cosh (0)

Simplify

ln (1) : 0

ln (1)

Apply log rule:

lo g_{a} (1) = 0

= 0

= 0 \cdot cosh (0)

Apply rule

0 \cdot a = 0

= 0

= 0 + tan (0)

0 + tan (0) = tan (0)

= tan (0)

= tan (0)

Use the following trivial identity:

tan (0) = 0

tan (0)

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 0

= 0

arctan (\frac{- 3}{- 3 3})

\frac{0.001}{sin ( 0.001 )}

cot (7 0^{\circ}) + 4 cos (7 0^{\circ})

- 8 cot (\frac{π}{2})

csc (arccot (3))

What is the value of [ln(0+1)]cosh(0)+tan(0) ?
The value of [ln(0+1)]cosh(0)+tan(0) is 0