Solution:
Given equation:
$\cot\text{x}-\tan\text{x}=\sec\text{x}$
$\Rightarrow\frac{\cos\text{x}}{\sin\text{x}}-\frac{\sin\text{x}}{\cos\text{x}}=\frac{1}{\cos\text{x}}$
$\Rightarrow\frac{\cos^2\text{x}-\sin^2}{\sin\text{x}\cosx}=\frac{1}{\cos\text{x}}$
$\Rightarrow\cos^2\text{x}\sin^2=\sin\text{x}$
$\Rightarrow(1-\sin^2\text{x})-\sin^2\text{x}=\sin\text{x}$
$\Rightarrow1-2\sin^2=\sin\text{x}$
$\Rightarrow2\sin^2\text{x}+\sin\text{x}-1=0$
$\Rightarrow2\sin^2\text{x}+2\sin\text{x}-\sin\text{x}-1=0$
$\Rightarrow2\sin\text{x}(\sin\text{x}+1)-1(\sin\text{x}+1)=0$
$\Rightarrow(\sin\text{x}+1)(2\sin\text{x}-1)=0$
$\Rightarrow\sin\text{x}+1=0$ or $2\sin\text{x}-1=0$
$\Rightarrow\sin\text{x}=-1$ or $\sin\text{x}=\frac{1}{2}$
Now,
$\sin\text{x}=\frac{1}{2}\Rightarrow\sin\text{x}=sin\frac{3\pi}{2}\Rightarrow\text{x}=\text{m}\pi+(-1)^\text{m}\frac{3\pi}{2},\ \text{m}\in\text{Z}$
And
$\sin\text{x}=\frac{1}{2}\Rightarrow\sin\text{x}=\sin\frac{\pi}{6}\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6},\text{n}\in\text{Z}$
$\therefore\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{6},\text{n}\in\text{Z}$
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The value of arg(x), when x < 0 is:
$0$
$\frac{\pi}{2}$
$\pi$
None of these