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COMEDK · Maths · 9. Trigonometric Equations

The general solution of \(\tan x-\sin x=1-\tan x \cdot \sin x\)

  1. A \(x=n \pi+\frac{\pi}{4}\) and or \(x=n \pi+(-1)^{n}\left(\frac{-\pi}{2}\right)\)
  2. B \(x=\frac{n \pi}{4}-\frac{\pi}{4}\) and or \(x=n \pi+(-1)^{n}\left(-\frac{\pi}{2}\right)\)
  3. C \(x=n \pi+\frac{\pi}{4}\)
  4. D \(x=n \pi+\frac{\pi}{6}\) and or \(\pi=n \pi+(-1)^{n}\left(-\frac{\pi}{2}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(x=n \pi+\frac{\pi}{4}\) and or \(x=n \pi+(-1)^{n}\left(\frac{-\pi}{2}\right)\)

Step-by-step Solution

Detailed explanation

\(\tan x-\sin x=1-\tan x \sin x\) \(\Rightarrow \tan x+\tan x \sin x=1+\sin x\) \(\tan x(1+\sin x)-(1+\sin x)=0\) \((1+\sin x)(\tan x-1)=0\) \(\Rightarrow \quad \tan x-1=0\) or \(1+\sin x=0\) \(\tan x=1=\tan \frac{\pi}{4} \quad\) or \(\sin x=-1=\sin \left(-\frac{\pi}{2}\right)\)…