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प्रश्न
If \[\sec \left( x + \alpha \right) + \sec \left( x - \alpha \right) = 2 \sec x\] , prove that \[\cos x = \pm \sqrt{2} \cos\frac{\alpha}{2}\]
उत्तर
Equation \[\sec \left( x + \alpha \right) + \sec \left( x - \alpha \right) = 2 \sec x\] can be written as \[\frac{1}{\cos\left( x + \alpha \right)} + \frac{1}{\cos\left( x - \alpha \right)} = \frac{2}{\text{ cos } x}\]
\[ \Rightarrow \frac{1}{\text{ cos } x \times cos\alpha - \text{ sin } x \times sin\alpha} + \frac{1}{\text{ cos } x \times cos\alpha + \text{ sin } x \times sin\alpha} = \frac{2}{\text{ cos } x} \left[ \because \cos\left( A + B \right) = \text{ cos } A \times \text{ cos } B - \text{ sin } A \times \text
{ sin } B \text{ and } \cos\left( A - B \right) = \text{ cos } A \times \text{ cos } B + \text{ sin } A \times \text{ sin } B \right] \]
\[ \Rightarrow \frac{2\text{ cos } x \times cos\alpha}{\cos^2 x \times \cos^2 \alpha - \sin^2 x \times \sin^2 \alpha} = \frac{2}{\text{ cos } x}\]
\[ \Rightarrow \frac{\text{ cos } x \times cos\alpha}{\cos^2 x \times \cos^2 \alpha - \left( 1 - \cos^2 x \right) \times \sin^2 \alpha} = \frac{1}{\text{ cos } x}\]
\[\Rightarrow \frac{\cos^2 x \times cos\alpha}{\cos^2 x \times \cos^2 \alpha - \left( 1 - \cos^2 x \right) \times \sin^2 \alpha} = 1\]
\[ \Rightarrow \frac{\cos^2 x \times cos\alpha}{\cos^2 x \times \cos^2 \alpha - \sin^2 \alpha + \cos^2 x \sin^2 \alpha} = 1\]
\[ \Rightarrow \cos^2 x \times cos\alpha = \cos^2 x \times \cos^2 \alpha - \sin^2 \alpha + \cos^2 x \sin^2 \alpha\]
\[ \Rightarrow \cos^2 x \times cos\alpha = \cos^2 x\left( \cos^2 \alpha + \sin^2 \alpha \right) - \sin^2 \alpha\]
\[ \Rightarrow \cos^2 x \times cos\alpha = \cos^2 x - \sin^2 \alpha\]
\[\Rightarrow \cos^2 x \times cos\alpha - \cos^2 x = - \sin^2 \alpha\]
\[ \Rightarrow \cos^2 x\left( cos\alpha - 1 \right) = - \sin^2 \alpha\]
\[ \Rightarrow \cos^2 x\left( 1 - cos\alpha \right) = \sin^2 \alpha\]
\[ \Rightarrow \cos^2 x = \frac{\sin^2 \alpha}{2 \sin^2 \frac{\alpha}{2}} \left( \because 2 \sin^2 \frac{x}{2} = 1 - \text{ cos } x \right)\]
\[\Rightarrow \cos^2 x = \frac{4 \sin^2 \frac{\alpha}{2} \times \cos^2 \frac{\alpha}{2}}{2 \sin^2 \frac{\alpha}{2}} \left( \because \sin^2 x = 4 \sin^2 \frac{x}{2} \times \cos^2 \frac{x}{2} \right) \]
\[ \Rightarrow \text{ cos } x = \pm \sqrt{2} \cos\frac{\alpha}{2}\]
\[\text{ Hence proved } .\]
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