What is the value of ln|tan(0)+sec(0)| ?

The value of ln|tan(0)+sec(0)| is 0

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ln|tan(0)+sec(0)|

ln ∣ tan (0) + sec (0) ∣

0

Solution steps

ln ∣ tan (0) + sec (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

Use the following trivial identity:

sec (0) = 1

sec (0)

sec (x)

periodicity table with

2 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

= 1

= ln ∣ 0 + 1 ∣

Simplify

ln ∣ 0 + 1 ∣ : 0

ln ∣ 0 + 1 ∣

Add the numbers:

0 + 1 = 1

= ln ∣ 1 ∣

Simplify

∣ 1 ∣ : 1

∣ 1 ∣

Apply absolute rule:

∣ a ∣ = a, a \geq 0

∣ 1 ∣ = 1

= 1

= ln (1)

Apply log rule:

lo g_{a} (1) = 0

= 0

= 0

arcsin (\frac{9}{13})

cos^{2} (\frac{π}{2}) - 2 sin (\frac{π}{2})

- \frac{1}{2} cos (π)

60 sin (2 0^{\circ})

- csc (0)