If \(\frac{\sin\spaceθ \space+\space \cos\spaceθ} {\sin \spaceθ \space-\space \cos \spaceθ} = \frac{\sqrt3 \space-\space 1}{\sqrt3 \space+\space 1} \) , then the angle θ is
Solving Trigonometric Equations: Finding Angle θ
The problem asks us to find the value of the angle \(\theta\) given a specific trigonometric equation relating \(\sin\theta\) and \(\cos\theta\).
The given equation is:
$$ \frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = \frac{\sqrt3 - 1}{\sqrt3 + 1} $$
To solve this equation for \(\theta\), we can use a property called Componendo and Dividendo. This rule states that if \(\frac{a}{b} = \frac{c}{d}\), then \(\frac{a+b}{a-b} = \frac{c+d}{c-d}\).
Let \(a = \sin\theta + \cos\theta\) and \(b = \sin\theta - \cos\theta\). Also, let \(c = \sqrt3 - 1\) and \(d = \sqrt3 + 1\).
Applying the Componendo and Dividendo rule to the given equation:
$$ \frac{(\sin\theta + \cos\theta) + (\sin\theta - \cos\theta)}{(\sin\theta + \cos\theta) - (\sin\theta - \cos\theta)} = \frac{(\sqrt3 - 1) + (\sqrt3 + 1)}{(\sqrt3 - 1) - (\sqrt3 + 1)} $$
Now, let's simplify both sides of the equation:
The numerator of the left side is \((\sin\theta + \cos\theta) + (\sin\theta - \cos\theta) = \sin\theta + \cos\theta + \sin\theta - \cos\theta = 2\sin\theta\).
The denominator of the left side is \((\sin\theta + \cos\theta) - (\sin\theta - \cos\theta) = \sin\theta + \cos\theta - \sin\theta + \cos\theta = 2\cos\theta\).
So, the left side simplifies to \(\frac{2\sin\theta}{2\cos\theta} = \frac{\sin\theta}{\cos\theta} = \tan\theta\).
The numerator of the right side is \((\sqrt3 - 1) + (\sqrt3 + 1) = \sqrt3 - 1 + \sqrt3 + 1 = 2\sqrt3\).
The denominator of the right side is \((\sqrt3 - 1) - (\sqrt3 + 1) = \sqrt3 - 1 - \sqrt3 - 1 = -2\).
So, the right side simplifies to \(\frac{2\sqrt3}{-2} = -\sqrt3\).
Equating the simplified left and right sides, we get:
$$ \tan\theta = -\sqrt3 $$
Now we need to find the angle \(\theta\) for which \(\tan\theta = -\sqrt3\). We know that \(\tan 60^\circ = \sqrt3\). Since the tangent is negative, \(\theta\) must be in the second or fourth quadrant.
- In the second quadrant, the angle is \(180^\circ - 60^\circ = 120^\circ\). \(\tan 120^\circ = -\tan 60^\circ = -\sqrt3\).
- In the fourth quadrant, the angle is \(360^\circ - 60^\circ = 300^\circ\). \(\tan 300^\circ = -\tan 60^\circ = -\sqrt3\).
The general solution for \(\tan\theta = \tan\alpha\) is \(\theta = n \cdot 180^\circ + \alpha\), where \(n\) is an integer and \(\alpha\) is a principal angle (like \(120^\circ\) or \(-60^\circ\)).
We are given options for \(\theta\). Let's check which option satisfies \(\tan\theta = -\sqrt3\):
- For \(\theta = 45^\circ\): \(\tan 45^\circ = 1\). This is not \(-\sqrt3\).
- For \(\theta = 90^\circ\): \(\tan 90^\circ\) is undefined.
- For \(\theta = 240^\circ\): \(\tan 240^\circ = \tan (180^\circ + 60^\circ) = \tan 60^\circ = \sqrt3\). This is not \(-\sqrt3\).
- For \(\theta = 300^\circ\): \(\tan 300^\circ = \tan (360^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt3\). This matches the required value.
Also, for the original equation to be defined, the denominator \(\sin\theta - \cos\theta\) must not be zero. This happens when \(\sin\theta = \cos\theta\), which means \(\tan\theta = 1\). This occurs at angles like \(45^\circ\), \(225^\circ\), etc. Since \(\tan 300^\circ = -\sqrt3 \neq 1\), the denominator is not zero for \(\theta = 300^\circ\).
Thus, the angle \(\theta\) that satisfies the equation is \(300^\circ\).
| Angle (\(\theta\)) | \(\tan\theta\) | Matches \(-\sqrt3\)? |
|---|---|---|
| \(45^\circ\) | \(1\) | No |
| \(90^\circ\) | Undefined | No |
| \(240^\circ\) | \(\sqrt3\) | No |
| \(300^\circ\) | \(-\sqrt3\) | Yes |
Revision Table: Key Trigonometric Values
| Angle | \(\sin\theta\) | \(\cos\theta\) | \(\tan\theta\) |
|---|---|---|---|
| \(60^\circ\) | \(\frac{\sqrt3}{2}\) | \(\frac{1}{2}\) | \(\sqrt3\) |
| \(120^\circ\) | \(\frac{\sqrt3}{2}\) | \(-\frac{1}{2}\) | \(-\sqrt3\) |
| \(300^\circ\) | \(-\frac{\sqrt3}{2}\) | \(\frac{1}{2}\) | \(-\sqrt3\) |
Additional Information: Componendo and Dividendo Rule
The Componendo and Dividendo rule is a useful algebraic technique derived from proportions. If you have a proportion \(\frac{a}{b} = \frac{c}{d}\), it implies that \(\frac{a+b}{b} = \frac{c+d}{d}\) (Componendo) and \(\frac{a-b}{b} = \frac{c-d}{d}\) (Dividendo). By dividing the Componendo result by the Dividendo result (assuming \(b \neq 0\) and \(d \neq 0\)), we get \(\frac{(a+b)/b}{(a-b)/b} = \frac{(c+d)/d}{(c-d)/d}\), which simplifies to \(\frac{a+b}{a-b} = \frac{c+d}{c-d}\). This rule is particularly helpful in simplifying equations involving fractions of sums and differences, as seen in this trigonometric problem.
Alternatively, one could solve the original equation by cross-multiplication:
\((\sin\theta + \cos\theta)(\sqrt3 + 1) = (\sin\theta - \cos\theta)(\sqrt3 - 1)\)
\(\sqrt3\sin\theta + \sin\theta + \sqrt3\cos\theta + \cos\theta = \sqrt3\sin\theta - \sin\theta - \sqrt3\cos\theta + \cos\theta\)
Rearranging terms to group \(\sin\theta\) and \(\cos\theta\):
\(\sin\theta + \sqrt3\cos\theta = -\sin\theta - \sqrt3\cos\theta\)
\(\sin\theta + \sin\theta = -\sqrt3\cos\theta - \sqrt3\cos\theta\)
\(2\sin\theta = -2\sqrt3\cos\theta\)
\(\sin\theta = -\sqrt3\cos\theta\)
If \(\cos\theta \neq 0\), we can divide by \(\cos\theta\):
\(\frac{\sin\theta}{\cos\theta} = -\sqrt3\)
\(\tan\theta = -\sqrt3\)
This leads to the same result and confirms the use of Componendo and Dividendo was valid and efficient.
Important Questions from Trigonometry
The given equation can be reduced to
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The value of 5 sin 14° sec 76° + 3 cot 15° cot 75° + 2 tan 45° is:
If two complimentary angles are in the ratio of 4 : 5, find the greater angle.
If tan α = 1/2, tan β = 1/3, then find α + β.
