Correct Answer - Option 3 :
\(\frac {7\pi}{12}, \frac {11\pi}{12}\)
Concept:
- \(\tan \theta = \frac{{\sin \theta }}{{\cos \theta }}\)
- \({\sin ^2}x + co{s^2}x = 1\)
- \(si{n^4}x + co{s^4}x = 1 - 2{\sin ^2}xco{s^2}x\)
Calculation:
Given: sin2 x tan x + cos2 x cot x - sin 2x = 1 + tan x + cot x
\(⇒ \frac{{si{n^3}x}}{{\cos x}} + \frac{{co{s^3}x}}{{sinx}} - sin2x = 1 + \frac{{sinx}}{{cosx}} + \frac{{cosx}}{{sinx}}\)
\(⇒ \frac{{{{\sin }^4}x + {{\cos }^4}x}}{{\sin x\cos x}} - sin2x = 1 + \frac{{si{n^2}x + {{\cos }^2}x}}{{\sin xcosx}}\)
\(⇒ \frac{{1 - 2{{\sin }^2}xco{s^2}x}}{{\sin x\cos x}} - sin2x = 1 + \frac{1}{{\sin xcosx}}\)
\(⇒ \frac{1}{{\sin x\cos x}} - \frac{{2{{\sin }^2}xco{s^2}x}}{{\sin x\cos x}} - sin2x = 1 + \frac{1}{{\sin xcosx}}\)
⇒ - 2 sin x cos x - sin 2x = 1
⇒ - 2 sin 2x = 1
⇒ sin 2x = - 1/2
\(\Rightarrow 2x = \frac{{7\pi }}{6},\frac{{11\pi }}{6}\)
\(\Rightarrow x = \frac{{7\pi }}{{12}},\frac{{11\pi }}{{12}}\)
Hence, option C is the correct answer.
