To convert the equation of the line from Cartesian coordinates to polar coordinates, we need to use the relationship between the two coordinate systems. In Cartesian coordinates, the line is given by:
\(y = x - 1\)
In polar coordinates, the relationships are:
Substitute these into the Cartesian equation:
\(r \sin \theta = r \cos \theta - 1\)
Rearranging the terms gives:
\(r \cos \theta - r \sin \theta = 1\)
Factoring out \(r\) from the left side:
\(r (\cos \theta - \sin \theta) = 1\)
This is the polar form of the line. Thus, the correct option is:
\(r(\cos\theta - \sin\theta) = 1\)
Let's verify why this is the correct option and others are not:
Thus, the correct answer is \(r(\cos\theta - \sin\theta) = 1\).
Consider the hyperbolic functions in Group – 1 and their definitions in Group - 2.
| Group - 1 | Group - 2 | ||
| P | $\tanh x$ | I | $\frac{e^x + e^{-x}}{e^x - e^{-x}}$ |
| Q | $\coth x$ | II | $\frac{2}{e^x + e^{-x}}$ |
| R | $\text{sech } x$ | III | $\frac{2}{e^x - e^{-x}}$ |
| S | $\text{cosech } x$ | IV | $\frac{e^x - e^{-x}}{e^x + e^{-x}}$ |
The correct combination is