What is the value of sqrt(1+tan^2(45)) ?

The value of sqrt(1+tan^2(45)) is square root of 2

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sqrt(1+tan^2(45))

1 + tan^{2} (4 5^{\circ})

2

Decimal

1.41421 \dots

Solution steps

1 + tan^{2} (4 5^{\circ})

Use the following trivial identity:

tan (4 5^{\circ}) = 1

tan (4 5^{\circ})

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

= 1 + 1^{2}

Simplify

= 2

0.099 (9.8) (cos (2 2^{\circ})) + 0.71 (cos (6 8^{\circ}))

cos (\frac{2 π}{15}) cos (\frac{π}{30}) - sin (\frac{2 π}{15}) sin (\frac{π}{30})

cos (65. 7^{\circ})

120 (tan (32. 3^{\circ}))

\frac{cos ( 9 0 ^{\circ} ) + cos ( 0 )}{2 cos ( 5 0 ^{\circ} )}

What is the value of sqrt(1+tan^2(45)) ?
The value of sqrt(1+tan^2(45)) is square root of 2