Solving for Tan y
We are given the trigonometric equation:
\(6\sin y + \cos y = \sqrt{9}\sin y\)
First, simplify the square root:
\(\sqrt{9} = 3\)
Substitute this back into the equation:
\(6\sin y + \cos y = 3\sin y\)
Rearranging the Equation
To find \(\tan y\), we need to isolate the terms involving \(\sin y\) and \(\cos y\). Subtract \(3\sin y\) from both sides:
\(6\sin y - 3\sin y + \cos y = 0\)
\(3\sin y + \cos y = 0\)
Now, move the \(\cos y\) term to the other side:
\(3\sin y = -\cos y\)
Calculating Tan y
The definition of the tangent function is \(\tan y = \frac{\sin y}{\cos y}\). To get this form, divide both sides of the equation \(3\sin y = -\cos y\) by \(\cos y\) (assuming \(\cos y \neq 0\)):
\(\frac{3\sin y}{\cos y} = \frac{-\cos y}{\cos y}\)
Using the identity \(\tan y = \frac{\sin y}{\cos y}\):
\(3\tan y = -1\)
Finally, divide by 3 to solve for \(\tan y\):
\(\tan y = -\frac{1}{3}\)
Thus, the value of \(\tan y\) is \(-\frac{1}{3}\).
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