# If sin2 x tan x + cos2 x cot x - sin 2x = 1 + tan x + cot x, x ϵ (0, π), then x

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If sin2 x tan x + cos2 x cot x - sin 2x = 1 + tan x + cot x, x ϵ (0, π), then x
1. $\frac {3\pi}{12}, \frac {5\pi}{12}$
2. $\frac {5\pi}{12}, \frac {7\pi}{12}$
3. $\frac {7\pi}{12}, \frac {11\pi}{12}$
4. $\frac {7\pi}{12}, \frac {9\pi}{12}$

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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.