SATHEE: Trigonometrical Equations 1 Question 26

Trigonometrical Equations 1 Question 26

26. If $e^{[(\sin^{2} x + \sin^{4} x + \sin^{6} x + \dots + \infty) \log_{e} 2]}$ satisfies the equation $x^{2} - 9 x + 8 = 0$ ,then the value of $g (x) = \frac{\cos x}{\cos x + \sin x}$ is.

(1991, 4M)

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Answer:

Correct Answer: 26. $\frac{\sqrt{3} - 1}{2}$

Solution:

$e x p^{[(\sin^{2} x + \sin^{4} x + \sin^{6} x + \dots + \infty) \log_{e} 2]}$

$\begin{aligned} = e^{\frac{\sin^{2} x}{1 - \sin^{2} x} \cdot \log_{e} 2} = e^{\log_{e} 2 \frac{\sin^{2} x}{\cos^{2} x}} \\ \Rightarrow 2^{\tan^{2} x} satisfies x^{2} - 9 x + 8 = 0 \\ \Rightarrow x = 1, 8 \\ ∴ 2^{\tan^{2} x} = 1 and 2^{\tan^{2} x} = 8 \\ \Rightarrow \tan^{2} x = 0 and \tan^{2} x = 3 \\ \Rightarrow x = n π and \tan^{2} x = \tan \frac{π^{2}}{3} \\ \Rightarrow x = n π and x = n π \pm \frac{π}{3} \end{aligned}$

Neglecting $x = n π$ as $0 < x < \frac{π}{2}$

$\begin{aligned} \Rightarrow x = \frac{π}{3} \in 0, \frac{π}{2} \\ ∴ \frac{\cos x}{\cos x + \sin x} = \frac{\frac{1}{2}}{\frac{1}{2} + \frac{\sqrt{3}}{2}} = \frac{1}{1 + \sqrt{3}} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \\ \Rightarrow \frac{\cos x}{\cos x + \sin x} = \frac{\sqrt{3} - 1}{2} \end{aligned}$

Trigonometrical Equations 1 Question 26

26. If $e^{[(\sin^{2} x + \sin^{4} x + \sin^{6} x + \dots + \infty) \log_{e} 2]}$ satisfies the equation $x^{2} - 9 x + 8 = 0$ ,then the value of $g (x) = \frac{\cos x}{\cos x + \sin x}$ is.

Answer:

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Trigonometrical Equations 1 Question 26

26. If e[(sin2⁡x+sin4⁡x+sin6⁡x+…+∞)loge⁡2] satisfies the equation x2−9x+8=0 ,then the value of g(x)=cos⁡xcos⁡x+sin⁡x is.

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26. If $e^{[(\sin^{2} x + \sin^{4} x + \sin^{6} x + \dots + \infty) \log_{e} 2]}$ satisfies the equation $x^{2} - 9 x + 8 = 0$ ,then the value of $g (x) = \frac{\cos x}{\cos x + \sin x}$ is.