What is the general solution for arctan(x)+arctan(2)=arctan(7) ?

The general solution for arctan(x)+arctan(2)=arctan(7) is x= 1/3

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arctan(x)+arctan(2)=arctan(7)

arctan (x) + arctan (2) = arctan (7)

x = \frac{1}{3}

Solution steps

arctan (x) + arctan (2) = arctan (7)

Move

arctan (2)

to the right side

arctan (x) + arctan (2) = arctan (7)

Subtract

arctan (2)

from both sides

arctan (x) + arctan (2) - arctan (2) = arctan (7) - arctan (2)

Simplify

arctan (x) = arctan (7) - arctan (2)

arctan (x) = arctan (7) - arctan (2)

Apply trig inverse properties

arctan (x) = arctan (7) - arctan (2)

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

x = tan (arctan (7) - arctan (2))

tan (arctan (7) - arctan (2)) = \frac{1}{3}

tan (arctan (7) - arctan (2))

Rewrite using trig identities:

arctan (7) - arctan (2) = arctan (\frac{1}{3})

arctan (7) - arctan (2)

Use the Sum to Product identity:

arctan (s) - arctan (t) = arctan (\frac{s - t}{1 + s t})

= arctan (\frac{7 - 2}{1 + 7 \cdot 2})

Simplify:

\frac{7 - 2}{1 + 7 \cdot 2} = \frac{1}{3}

\frac{7 - 2}{1 + 7 \cdot 2}

Subtract the numbers:

7 - 2 = 5

= \frac{5}{1 + 7 \cdot 2}

1 + 7 \cdot 2 = 15

1 + 7 \cdot 2

Multiply the numbers:

7 \cdot 2 = 14

= 1 + 14

Add the numbers:

1 + 14 = 15

= 15

= \frac{5}{15}

Cancel the common factor:

5

= \frac{1}{3}

= arctan (\frac{1}{3})

= tan (arctan (\frac{1}{3}))

Rewrite using trig identities:

tan (arctan (\frac{1}{3})) = \frac{1}{3}

Use the following identity:

tan (arctan (x)) = x

= \frac{1}{3}

= \frac{1}{3}

x = \frac{1}{3}

x = \frac{1}{3}

3 sec (x) - 1 = 2

sec (x) = 6

sin (θ) = \frac{8}{12}

y, \int x d x = csc^{2} (\frac{y}{2})

sec (x) + 2 = 0, - π \leq x \leq \frac{3 π}{2}

What is the general solution for arctan(x)+arctan(2)=arctan(7) ?
The general solution for arctan(x)+arctan(2)=arctan(7) is x= 1/3