Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

If a _ 1 , a _ 2 , a _ 3 are in arithmetic progression and d is the...

If a1,a2,a3{a}_{1}, {a}_{2}, {a}_{3} are in arithmetic progression and dd is the common difference, then tan⁡−1(d1+a1a2)+tan⁡−1(d1+a2a3)=\tan^{-1} \left(\dfrac{d}{1 + {a}_{1}{a}_{2}}\right) + \tan^{-1} \left(\dfrac{d}{1 + {a}_{2}{a}_{3}}\right) =

Options

  1. Atan⁡−1(2d1+a1a3)\tan^{- 1} \left(\dfrac{2d}{1 + {a}_{1}{a}_{3}}\right)
  2. Btan⁡−1(d1+a1a3)\tan^{- 1} \left(\dfrac{d}{1 + {a}_{1}{a}_{3}}\right)
  3. Ctan⁡−1(2d1+a2a3)\tan^{- 1} \left(\dfrac{2d}{1 + {a}_{2}{a}_{3}}\right)
  4. Dtan⁡−1(2d1−a1a3)\tan^{- 1} \left(\dfrac{2d}{1 - {a}_{1}{a}_{3}}\right)

Correct answer

A. tan⁡−1(2d1+a1a3)\tan^{- 1} \left(\dfrac{2d}{1 + {a}_{1}{a}_{3}}\right)

Step-by-step solution

Since, a1,a2,a3{a}_{1}, {a}_{2}, {a}_{3} are in A.P.

⇒a2−a1=d=a3−a2⇒ {a}_{2} - {a}_{1} = d = {a}_{3} - {a}_{2}

Now, tan⁡−1(d1+a1a2)+tan⁡−1\tan^{-1} \left(\dfrac{d}{1 + {a}_{1}{a}_{2}}\right) + \tan^{-1} \left(\dfrac{d}{1 + {a}_{2}{a}_{3}}\right)<br/><br/>br /><br />= \tan^{-1} \left(\dfrac{{a}_{2} - {a}_{1}}{1 + {a}_{1}{a}_{2}}\right) + \tan^{-1}(a3−a21+a2a3)\left(\dfrac{{a}_{3} - {a}_{2}}{1 + {a}_{2}{a}_{3}}\right)

\left(∵ \tan^{-1}x - \tan^{-1}y = \tan^{-1}\dfrac{x - y}{1 + xy}\right)<br/><br/><br /><br />= \tan^{-1} {a}_{2} - \tan^{-1} {a}_{1} + \tan^{-1} {a}_{3} - \tan^{-1} {a}_{2}<br/><br/><br /><br />{= tan}^{- 1} {a}_{3} - \tan^{-1} {a}_{1}<br/><br/><br /><br />= \tan^{-1} \left(\dfrac{{a}_{3} - {a}_{1}}{1 + {a}_{1}{a}_{3}}\right)

=\tan^{-1}((a3−a2)+(a2−a1)1+a1a3)\left(\dfrac{ \left({a}_{3} - {a}_{2}\right) + \left({a}_{2} - {a}_{1}\right)}{1 + {a}_{1}{a}_{3}}\right)

=\tan^{-1}(2d1+a1a3)\left(\dfrac{2d}{1 + {a}_{1}{a}_{3}}\right)
Practice Inverse Trigonometric Functions on Quantrex Academy →

More from Inverse Trigonometric Functions

Considering only the principal values of the inverse trigonometric functions, the value… JEE Advanced 2026Let α = 3 ^ -1 ((6)/(11)) and β = 3 ^ -1 ((4)/(9)) , where inverse trigonometric… JEE Main 2026If ( ^ -1 (x 2 )) = ( ^ -1 1-x^2 ) , x ∈ (0,1) , then the value of x is : JEE Main 2026Let [·] denote the greatest integer function. If the domain of the function f(x) = ^ -1… JEE Main 2026Let 0 < α < 1 , β = (1)/(3α) and ^ -1 (1-α) + ^ -1 (1-β) = (π)/(4) . Then 6(α + β) is… JEE Main 2026If (π)/(4) + ∑_ p=1 ^ 11 ^ -1 ( 2^ p-1 1 + 2^ 2p-1 ) = α , then α is equal to __________. JEE Main 2026If y = ^ -1 ((3 x - 4 x)/(4 x + 3 x)) + 2 ^ -1 ( x 1+ 1-x^2 ) , then (dy)/(dx) at x = 3 2… JEE Main 2026Considering the principal values of inverse trigonometric functions, the value of the… JEE Main 2026If k= ((π)/(4)+(1)/(2) ^ -1 ((2)/(3)))+ ((1)/(2) ^ -1 ((2)/(3))) , then the number of… JEE Main 2026If the domain of the function f(x)= ^ -1 ( 1 x^ 2 -2 x-2 ) , is (-∞, α] [β, γ] [δ, ∞)… JEE Main 2026The number of solutions of ^ -1 4x + ^ -1 6x = (π)/(6) , where - 1 2 6 < x < 1 2 6 , is… JEE Main 2026Let the maximum value of ( ^ -1 x)^ 2 +( ^ -1 x)^ 2 for x ∈[- 3 2 , 1 2 ] be m n π^ 2… JEE Main 2026If the domain of the function f(x)= ^ -1 ((2 x-5)/(11-3 x))+ ^ -1 (2 x^ 2 -3 x+1) is the… JEE Main 2026The total number of real solutions of the equation θ= ^ -1 (2 θ)-(1)/(2) ^ -1 ((6 θ)/(9+… JEE Advanced 2025The value of ^ -1 ( 1+ ^2(2) -1 (2) )- ^ -1 ( 1+ ^2((1)/(2)) +1 ((1)/(2)) ) is equal to JEE Main 2025The sum of the infinite series ^ -1 ((7)/(4))+ ^ -1 ((19)/(4))+ ^ -1 ((39)/(4))+ ^ -1… JEE Main 2025 Full Inverse Trigonometric Functions list Inverse Trigonometric Functions JEE Maths PYQsJEE Mathematics PYQs All NTA Abhyas JEE Main PYQs