Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTRIGONOMETRIC FUNCTIONS4 Marks
Question
Find the general solution for each of the following equations: $\sec^22\text{x}=1-\tan2\text{x}$
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Answer
$\sec^22\text{x}=1-\tan2\text{x}$$\Rightarrow1+\tan^22\text{x}=1-\tan2\text{x}$
$\Rightarrow\tan^22\text{x}+\tan2\text{x}=0$
$\Rightarrow\tan2\text{x}(\tan2\text{x}+1)=0$
$\Rightarrow\tan2\text{x}=0$ or $\tan2\text{x}+1=0$
Now, $\tan2\text{x}=0$
$\Rightarrow\tan2\text{x}=\tan0$
$\Rightarrow2\text{x}=\text{n}\pi+0,$ where $\text{n}\in\text{Z}$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{2},$ where $\text{n}\in\text{Z}$
$\tan2\text{x}+1=0$
$\Rightarrow\tan2\text{x}=-1=-\tan\frac{\pi}{4}=\tan\Big(\pi-\frac{\pi}{4}\Big)=\tan\frac{3\pi}{4}$
$\Rightarrow2\text{x}=\text{n}\pi+\frac{3\pi}{8},$ where $\text{n}\in\text{Z}$
$\Rightarrow2\text{x}=\frac{\text{n}\pi}{2}+\frac{3\pi}{8}$ where $\text{n}\in\text{Z}$
Therefore, the general solution is $\frac{\text{n}\pi}{2}$ or $\frac{\text{n}\pi}{2}+\frac{3\pi}{8},\text{n}\in\text{Z}$
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