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

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

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

ln ∣ sec (0) + tan (0) ∣

0

Solution steps

ln ∣ sec (0) + tan (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

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

= ln ∣ 1 + 0 ∣

Simplify

ln ∣ 1 + 0 ∣ : 0

ln ∣ 1 + 0 ∣

Add the numbers:

1 + 0 = 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

cos (- \frac{1}{5})

arctan (\frac{7}{- 6})

\frac{sin ( 15 0 ^{\circ} ) + cos ( 30 0 ^{\circ} )}{tan ( 22 5 ^{\circ} ) - sin ( 30 0 ^{\circ} )}

26 cos (5 4^{\circ})

2 + sin (\frac{π}{6})