If u = cot^-1{√(tan θ)} - tan^-1{√(tan θ)} then, tan(π/4 - u/2) = A. √(tan θ ) B. √(cot θ) C. tan θ D. cot θ

Question

If u = cot^-1{\(\sqrt{tan\,\theta}\)} - tan^-1{\(\sqrt{tan\,\theta}\)} then, tan\((\frac{\pi}4-\frac{u}2)\) =

A. \(\sqrt{tan\,\theta}\)

B. \(\sqrt{cot\,\theta}\)

C. tan θ

D. cot θ

Answer 1

Correct option is A.\(\sqrt{tan\,\theta}\)

We are given with

u = cot ^-1{√tan θ} – tan^-1{√tan θ}

We need to find the value of \(tan(\frac{\pi}4-\frac{u}2).\)

Let √tan θ = x

Then, u = cot^-1{√tan θ} – tan^-1{√tan θ} can be written as

u = cot^-1 x – tan^-1 x …(i)

We know by the property of inverse trigonometry,

cot^-1 x – tan^-1 x = \(\frac \pi2\)

Or,

cot^-1 x = \(\frac \pi2\) - tan^-1 x

Substituting the value of cot^-1 x in equation (i), we get

Now, divide by 2 on both sides of the equation.