Question

Solve for x: 2 tan⁻¹`(1/3)` + sec⁻¹`((5sqrt2)/7)` = tan ⁻¹ x

Answer 1

Step 1: Simplify the first term

Using the formula: `2 tan^(−1) A = tan−1 ((2A)/(1 − A^2))`

`2 tan^(−1)(1/3) = tan^(−1) ((2/3)/(1 − 1/9)) `

`= tan^(−1) ((2/3)/(8/9)) = tan^(−1)(3/4)`

Step 2: Convert the second term to tan⁻¹

Let sec⁻¹`((5sqrt2)/7) = θ ⇒ sec θ = (5sqrt2)/7`

Using the identity tan θ = `sqrt(sec^2 θ − 1)`

tan θ = `sqrt(((5sqrt2)/7)^2) −1`

`= sqrt50/49 − 1`

`= sqrt1/49 = 1/7`

so, `sec−1 ((5sqrt2)/7) = tan^(−1)(1/7)`

Step 3: Combine and solve for 

Substitute these back into the original equation:

`= tan^(−1) (3/4) + tan^(−1) (1/7) = tan^(−1) x`

Using the formula `tan^(−1) A + tan^(−1) B = tan^(−1) ((A + B)/(1 − AB))`

`tan^(−1) ((3/4 + 1/7)/(1 − (3/4 × 1/7))) = tan^(−1) x`

`tan^(−1) ((25/28)/(25/28)) = tan^(−1) x`

`tan^(−1) (1) = tan^(−1) x`

⇒ x = 1