What is the value of ln|tan(pi/4)+sec(pi/4)| ?

The value of ln|tan(pi/4)+sec(pi/4)| is ln(1+sqrt(2))

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ln|tan(pi/4)+sec(pi/4)|

ln tan (\frac{π}{4}) + sec (\frac{π}{4})

ln (1 + 2)

Decimal Notation

0.88137 \dots

Solution steps

ln tan (\frac{π}{4}) + sec (\frac{π}{4})

Use the following trivial identity:

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

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}

= 1

Use the following trivial identity:

sec (\frac{π}{4}) = 2

sec (\frac{π}{4})

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}

= 2

= ln 1 + 2

Simplify

ln 1 + 2 : ln (1 + 2)

ln 1 + 2

Simplify

1 + 2 : 1 + 2

1 + 2

Apply absolute rule:

∣ A ∣ = A, A \geq 0

1 + 2 = 1 + 2

= 1 + 2

= ln (1 + 2)

= ln (1 + 2)

6 \cdot sin (6 0^{\circ})

tan (arctan (- 2))

arccos (\frac{12}{14})

20 sin (2 0^{\circ})

arctan (\frac{- 6}{2})

What is the value of ln|tan(pi/4)+sec(pi/4)| ?
The value of ln|tan(pi/4)+sec(pi/4)| is ln(1+sqrt(2))