\(\rm tan^{-1}x+cot^{-1}x=\frac{\pi}{2}\) लागू होता है, जब ____ है।

This question was previously asked in
NDA (Held On: 18 Apr 2021) Maths Previous Year paper
View all NDA Papers >
  1. x ∈ R
  2. केवल x ∈ R - (-1, 1) 
  3. केवल x ∈ R - {0} 
  4. केवल x ∈ R - [-1, 1] 

Answer (Detailed Solution Below)

Option 1 : x ∈ R
Free
NDA 01/2025: English Subject Test
5.5 K Users
30 Questions 120 Marks 30 Mins

Detailed Solution

Download Solution PDF

अवधारणा:

\(\rm tan^{-1}(x)+tan^{-1}({y})\) = \(\rm tan^{-1}\frac{x+y}{1-x\times y}\)

\(\rm cot^{-1}x=\) \(\rm tan^{-1}({1\over x})\)

गणना:

दिया गया है, \(\rm tan^{-1}x+cot^{-1}x=\frac{\pi}{2}\)

⇒ \(\rm tan^{-1}(x)+tan^{-1}({1\over x})=\frac{\pi}{2}\)

⇒ \(\rm tan^{-1}(x)+tan^{-1}({1\over x})=\frac{\pi}{2}\)

⇒ \(\rm tan^{-1}\frac{x+\frac{1}{x}}{1-x\times \frac{1}{x}} = \frac{\pi}{2}\)

⇒ \(\rm tan^{-1}\frac{x+\frac{1}{x}}{0} = \frac{\pi}{2}\)

⇒ \(\rm tan^{-1}\frac{x+\frac{1}{x}}{0} = \frac{\pi}{2}\)

⇒ \(\rm tan^{-1}({\infty}) = \frac{\pi}{2}\)

यह सभी x ∈ R के लिए सत्य है

Latest NDA Updates

Last updated on Jun 18, 2025

->UPSC has extended the UPSC NDA 2 Registration Date till 20th June 2025.

-> A total of 406 vacancies have been announced for NDA 2 Exam 2025.

->The NDA exam date 2025 has been announced. The written examination will be held on 14th September 2025.

-> The selection process for the NDA exam includes a Written Exam and SSB Interview.

-> Candidates who get successful selection under UPSC NDA will get a salary range between Rs. 15,600 to Rs. 39,100. 

-> Candidates must go through the NDA previous year question paper. Attempting the NDA mock test is also essential. 

More Inverse Trigonometric Functions Questions

Get Free Access Now
Hot Links: teen patti all app all teen patti master teen patti rules