যদি \(\tan x=\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta} \) , \(\frac{\pi}{4}<\theta<\frac {\pi}{2} \) হয়, তাহলে √2 sin x এর মান কত?

This question was previously asked in
CDS Elementary Mathematics 3 Sep 2023 Official Paper
View all CDS Papers >
  1. sin θ + cos θ
  2. sin θ - cos θ
  3. \(\frac{\sin \theta+\cos \theta}{2}\)
  4. \(\frac{\sin \theta-\cos \theta}{2}\)

Answer (Detailed Solution Below)

Option 1 : sin θ + cos θ
Free
UPSC CDS 01/2025 General Knowledge Full Mock Test
8.1 K Users
120 Questions 100 Marks 120 Mins

Detailed Solution

Download Solution PDF

প্রদত্ত:

tan x = \(\frac{\sin θ+\cos θ}{\sin θ-\cos θ}\) , \(\frac{\pi}{4}<θ<\frac{\pi}{2 } \)

অনুসৃত সূত্র:

1 tan θ = \(\frac{P}{B}\)

2 sin θ = \(\frac{P}{H}\)

3 sin2θ + cos2θ = 1

গণনা:

F1 Vinanti Defence 01.12.23 D1

পিথাগোরাসের উপপাদ্য দ্বারা,

⇒ H = \(\sqrt{(sinθ+cosθ)^2+(sinθ-cosθ)^2} \)

⇒ H = \(\sqrt{sin^2θ+cos^2θ+2sinθ cosθ+sin^2θ+cos^2θ-2sinθ cosθ} \)

⇒ H = \(\sqrt{2(sin^2θ+cos^2θ)} \) = \(\sqrt{2} \)

চিত্র অনুযায়ী,

\(\sqrt{2} \) sin x = \(\sqrt{2} \) x \(\frac{sinθ+\cosθ}{\sqrt{2}}\) = sinθ + cosθ

\(\sqrt{2} \) sin x এর নির্ণেয় মান হল sinθ + cosθ

Latest CDS Updates

Last updated on Jun 26, 2025

-> The UPSC CDS Exam Date 2025 has been released which will be conducted on 14th September 2025.

-> Candidates had applied online till 20th June 2025.

-> The selection process includes Written Examination, SSB Interview, Document Verification, and Medical Examination.  

-> Attempt UPSC CDS Free Mock Test to boost your score.

-> Refer to the CDS Previous Year Papers to enhance your preparation. 

Get Free Access Now
Hot Links: dhani teen patti teen patti wink teen patti master official teen patti boss teen patti win