Question

`50tan(3tan^-1(1/2) + 2cos^-1(1/sqrt(5))) + 4sqrt(2) tan(1/2tan^-1(2sqrt(2)))` is equal to ______.

Options

• 26

• 28

• 29

• 30

Answer 1

`50tan(3tan^-1(1/2) + 2cos^-1(1/sqrt(5))) + 4sqrt(2) tan(1/2tan^-1(2sqrt(2)))` is equal to 29.

Explanation:

Let A = 50 tan

`(3tan^-1(1/2) + 2cos^-1(1/sqrt(5))) + 4sqrt(2) tan(1/2tan^-1(2sqrt(2)))`

Let B = `tan(1/2tan^-1(2sqrt(2)))`

Let 2θ = `tan^-1(2sqrt(2));2θ ∈ (- π/2, π/2)`

∴ tanθ = `2sqrt(2)`

⇒ `(2tanθ)/(1 - tan^2θ) = 2sqrt(2)`  ...(i)

⇒ `2sqrt(2) tan^2θ + 2tanθ - 2sqrt(2)` = 0

⇒ `2sqrt(2) tan^2θ + 4tanθ - 2tanθ - 2sqrt(2)` = 0

⇒ `(tanθ + sqrt(2))(2sqrt(2)tanθ - 2)` = 0

⇒ `tanθ = - sqrt(2)` or `1/sqrt(2)`

∵ `θ ∈ (-π/4, π/4)`

∴ tanθ = `-sqrt(2)` is not possible

So, tanθ = `1/sqrt(2)`

∴ B = `1/sqrt(2)`

Now, `cos^-1(1/sqrt(5))` = tan^–1(2)

Let C = `tan(3tan^-1(1/2) + 2cos^-1(1/sqrt(5)))`

⇒ C = `tan(tan^-1(1/2) + 2tan^-1(1/2) + 2tan^-1(2))`

⇒ C = `tan(tan^-1(1/2) + 2[tan^-1((1/2 + 2)/(1 - 1))])`

⇒ C = `tan(tan^-1(1/2) + 2(π/2))`

⇒ C = `(π + tan^-1(1/2))`

⇒ C = `tan(tan^-1(1/2))`

⇒ C = `1/2`

∴ A = `50(1/2) + 4sqrt(2)(1/sqrt(2))`

⇒ A = 25 + 4 = 29