যদি α এবং β ধনাত্মক কোণ হয় যেমন \(α + β = \dfrac{\pi}{4}\) , তাহলে (1 + tan α) (1 + tan β) এর মান কত?

  1. 0
  2. 1
  3. 2
  4. 3

Answer (Detailed Solution Below)

Option 3 : 2
Free
NDA 01/2025: English Subject Test
5.5 K Users
30 Questions 120 Marks 30 Mins

Detailed Solution

Download Solution PDF

ধারণা:

\(\rm \tan (α + β) = \dfrac {tan \alpha + tan \beta }{1 -tan \alpha \;tan \beta }\)

 

গণনা:

প্রদত্ত, α এবং β হল ধনাত্মক কোণ যেমন \(α + β = \dfrac{\pi}{4}\) ,

\(\rm \tan (α + β) = \tan (\dfrac{\pi}{4})\)

\(\rm \dfrac {tan \alpha + tan \beta }{1 -tan \alpha \;tan \beta } = 1\)

\(\rm {tan \alpha + tan \beta }= 1 -tan \alpha \;tan \beta \)

\(\rm {tan \alpha +tan \alpha \;tan \beta + tan \beta } - 1 = 0 \)

\(\rm {tan \alpha +tan \alpha \;tan \beta + tan \beta } +1-2= 0 \)

\(\rm {tan \alpha +tan \alpha \;tan \beta + tan \beta } +1=2\)

\(\rm tan \alpha (1 +tan \beta ) + (1+tan \beta)=2\)

\(\rm (1+ tan \alpha )(1 +tan \beta )=2\)

সুতরাং, α এবং β যদি ধনাত্মক কোণ হয় যেমন \(α + β = \dfrac{\pi}{4}\) , তাহলে (1 + tan α) (1 + tan β) 2 এর সমান।

Latest NDA Updates

Last updated on May 30, 2025

->UPSC has released UPSC NDA 2 Notification on 28th May 2025 announcing the NDA 2 vacancies.

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

->The NDA exam date 2025 has been announced for cycle 2. The written examination will be held on 14th September 2025.

-> Earlier, the UPSC NDA 1 Exam Result has been released on the official website.

-> 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. 

Get Free Access Now
Hot Links: teen patti diya teen patti online game online teen patti teen patti vungo