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Question

# The most general solutions of the equation $secx-1=\left(\sqrt{2}-1\right)\mathrm{tan}x$ are given by

A

$n\mathrm{\pi }+\frac{\mathrm{\pi }}{8}$

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B

$2n\mathrm{\pi },2\mathrm{n\pi }+\frac{\mathrm{\pi }}{4}$

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C

$2n\mathrm{\pi }$

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D

None of these

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Solution

## The correct option is B $2n\mathrm{\pi },2\mathrm{n\pi }+\frac{\mathrm{\pi }}{4}$Explanation for the correct option:Find the general solution of the given equationGiven equation, $secx-1=\left(\sqrt{2}-1\right)\mathrm{tan}x...\left(1\right)$ We know that, $se{c}^{2}x-1={\mathrm{tan}}^{2}x$.So,$\left(secx-1\right)\left(secx+1\right)={\mathrm{tan}}^{2}x\phantom{\rule{0ex}{0ex}}⇒\left(secx+1\right)=\frac{{\mathrm{tan}}^{2}x}{secx-1}$From equation $\left(1\right)$.$\left(sex+1\right)=\frac{{\mathrm{tan}}^{2}x}{\left(\sqrt{2}-1\right)\mathrm{tan}x}\phantom{\rule{0ex}{0ex}}⇒\left(secx+1\right)=\frac{\mathrm{tan}x}{\left(\sqrt{2}-1\right)}\phantom{\rule{0ex}{0ex}}⇒\left(secx+1\right)=\frac{\mathrm{tan}x}{\left(\sqrt{2}-1\right)}·\frac{\left(\sqrt{2}+1\right)}{\left(\sqrt{2}+1\right)}\phantom{\rule{0ex}{0ex}}⇒\left(secx+1\right)=\left(\sqrt{2}+1\right)\mathrm{tan}x...\left(2\right)$Add equation $\left(1\right)$ and equation $\left(2\right)$.$2secx=2\sqrt{2}\mathrm{tan}x\phantom{\rule{0ex}{0ex}}⇒\frac{1}{\mathrm{cos}x}=\sqrt{2}\frac{\mathrm{sin}x}{\mathrm{cos}x}$Which is possible when, $\mathrm{sin}x=\frac{1}{\sqrt{2}}$ and $\mathrm{cos}x=1$.Therefore, The most general solutions of the equation $secx-1=\left(\sqrt{2}-1\right)\mathrm{tan}x$ are given by $2n\mathrm{\pi },2\mathrm{n\pi }+\frac{\mathrm{\pi }}{4}$.Hence, the correct option is $B$.

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