Prove that
\$4 (2 "tan"^-1(1/3) + "tan"^-1 (1/7)) = pi\$.

The given function is

\$4 (2 "tan"^-1(1/3) + "tan"^-1 (1/7)) = pi\$

Let’s start by letting:

\$α = tan^-1 (1/3)\$ and \$β = tan^-1 (1/7)\$

Since \$α = tan^−1(1/3​)\$, we have \$tanα =1/3​\$

The given equation can be rewritten using α and β: 4(2α + β) = π.

Calculate tan⁡(2α) using the double-angle formula for tangent:

\$tan⁡(2α) = (tan⁡(2α)) / (1- tan⁡^2(α))\$

Plug the values in the formula and then simplify it

\$"tan"⁡(2α) = 2 times (1/(3))/(1 - (1/3)^2)\$

\$"tan"⁡(2α) = (2/(3))/(1 - 1/(9))\$

\$"tan"⁡(2α) = (2/(3))/(8/(9))\$

\$"tan"⁡(2α) = 3/4\$

The tangent of a sum of two angles is given by:

\$"tan"("A" + "B") = ("tanA" + "tanB")/(1 − "tanA tanB")\$

The formula is rewritten as calculate tan(2α + β) using the tangent addition:

\$"tan"(2α + β) = ("tan"(2α) + "tan"(β))/(1 − "tan"(2α)"tan"(β))\$

Plug the values in the formula and then simplify it

\$"tan"(2α + β) = (3/(4) + 1/(7))/(1 - 3/(4) . 1/(7))\$

\$"tan"(2α + β) = (21/(28) + 4/(28))/(1 - 3/(28))\$

\$"tan"(2α + β) = (25/(28))/(25/(28))\$

\$"tan"(2α + β) = 1\$

Since tan⁡(2α + β )= 1 , we know:

\$(2α + β) = "tan"^-1(1)\$

\$(2α + β) = (pi)/4\$

Thus, substituting back, we get:

\$4(2α + β) = 4 times (pi)/4 = (pi)\$

Therefore, the given equation: \$4 (2 "tan"^-1(1/3) + "tan"^-1 (1/7) = (pi)\$ is proven to be true.

This is incorrect because the statement we needed to prove does not result in the value 1; it results in π

1

Option B is correct because the equation \$4(2"tan"^⁡−1(1/3) + "tan"⁡^−1(1/7)) = π\$ is proven to be true by using the formulas

Proved

Wrong: Because the equation does not equal zero

Zero

The "Not proved" option is incorrect because we have shown that the original equation is true using trigonometric identities and properties of the arctangent function

Not proved

Option

B

Can you summarize what you’ve understood in the above steps?