What is the value of sqrt(1+tan^2(pi/4)) ?

The value of sqrt(1+tan^2(pi/4)) is square root of 2

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sqrt(1+tan^2(pi/4))

1 + tan^{2} (\frac{π}{4})

2

Decimal Notation

1.41421 \dots

Solution steps

1 + tan^{2} (\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

= 1 + 1^{2}

Simplify

= 2

arctan (\frac{2.5}{4})

tan (\frac{π}{3}) - tan (\frac{π}{6})

cos (\frac{2 π}{3}) - arctan (1)

cot (arccos (- \frac{2}{5}))

40 sin (7 0^{\circ})

What is the value of sqrt(1+tan^2(pi/4)) ?
The value of sqrt(1+tan^2(pi/4)) is square root of 2