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(B) The intercept form of a line is given by $\frac{x}{a'} + \frac{y}{b'} = 1$,where $a'$ and $b'$ are the intercepts on the $x$ and $y$ axes respecti

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$x \cos 	heta + y \sin 	heta - 2a = 0$

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(B) The intercept form of a line is given by $\frac{x}{a'} + \frac{y}{b'} = 1$,where $a'$ and $b'$ are the intercepts on the $x$ and $y$ axes respectively.Given intercepts are $a' = 2a \sec 	heta$ and $b' = 2a \csc 	heta$.Substituting these into the intercept form:$\frac{x}{2a \sec 	heta} + \frac{y}{2a \csc 	heta} = 1$Since $\frac{1}{\sec 	heta} = \cos 	heta$ and $\frac{1}{\csc 	heta} = \sin 	heta$,we get:$\frac{x \cos 	heta}{2a} + \frac{y \sin 	heta}{2a} = 1$Multiplying both sides by $2a$:$x \cos 	heta + y \sin 	heta = 2a$Rearranging the terms,we get:$x \cos 	heta + y \sin 	heta - 2a = 0$.

The equation of the line which cuts off the intercepts $2a \sec \theta$ and $2a \csc \theta$ on the axes is

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