What is the general solution for arctan(t)+1=1 ?

The general solution for arctan(t)+1=1 is t=0

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arctan(t)+1=1

arctan (t) + 1 = 1

t = 0

Solution steps

arctan (t) + 1 = 1

Subtract

1

from both sides

arctan (t) + 1 - 1 = 1 - 1

Simplify

arctan (t) = 0

Apply trig inverse properties

arctan (t) = 0

arctan (x) = a \Rightarrow x = tan (a)

t = tan (0)

tan (0) = 0

tan (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

= 0

t = 0

t = 0

2 cos^{2} (θ) - 7 cos (θ) + 3 = 0, 0 < θ < 2 π

csc (θ) = \frac{3}{2}

12 cos^{2} (x) + 5 sin (x) - 9 = 0

2 + 2 cos (θ) = 2

1 - sec^{2} (x) = 0

What is the general solution for arctan(t)+1=1 ?
The general solution for arctan(t)+1=1 is t=0