For the following two (02) items: Let $p\sin^2 \alpha+q\cos^2 \alpha=m$, $q\sin^2 \beta+p\cos^2\beta=n$; $p \neq m, n$ and $q \neq m, n$.
Solving for the Ratio of Tangents
We are given two trigonometric equations:
- Equation 1: \(p\sin^2 \alpha+q\cos^2 \alpha=m\)
- Equation 2: \(q\sin^2 \beta+p\cos^2\beta=n\)
Our objective is to determine the value of the ratio \(\frac{\tan^2 \alpha}{\tan^2 \beta}\). To do this, we first need to find individual expressions for \(\tan^2 \alpha\) and \(\tan^2 \beta\) from the given equations.
Deriving \(\\\tan^2 \alpha\)
Let's start with Equation 1: \(p\sin^2 \alpha+q\cos^2 \alpha=m\).
To express this equation in terms of \(\tan^2 \alpha\), we can divide every term by \(\cos^2 \alpha\). We know \(\cos^2 \alpha\) cannot be zero because if it were, \(\sin^2 \alpha\) would be 1, leading to \(p(1) + q(0) = m\), which means \(p=m\). However, the problem states \(p \neq m\). So, division by \(\cos^2 \alpha\) is valid.
\(p \frac{\sin^2 \alpha}{\cos^2 \alpha} + q \frac{\cos^2 \alpha}{\cos^2 \alpha} = \frac{m}{\cos^2 \alpha}\)
Using the trigonometric identities \(\frac{\sin \theta}{\cos \theta} = \tan \theta\) and \(\frac{1}{\cos \theta} = \sec \theta\), and the identity \(\sec^2 \theta = 1 + \tan^2 \theta\), we can rewrite the equation:
\(p\tan^2 \alpha + q = m\sec^2 \alpha\)
Substitute \(\sec^2 \alpha = 1 + \tan^2 \alpha\):
\(p\tan^2 \alpha + q = m(1 + \tan^2 \alpha)\)
Expand the right side:
\(p\tan^2 \alpha + q = m + m\tan^2 \alpha\)
Now, we rearrange the terms to group the \(\tan^2 \alpha\) terms on one side and the constants on the other:
\(p\tan^2 \alpha - m\tan^2 \alpha = m - q\)
Factor out \(\tan^2 \alpha\) from the terms on the left:
\((p - m)\tan^2 \alpha = m - q\)
Since \(p \neq m\), we can divide both sides by \((p - m)\) to find \(\tan^2 \alpha\):
\(\tan^2 \alpha = \frac{m-q}{p-m}\)
Deriving \(\\\tan^2 \beta\)
Next, we apply a similar process to Equation 2: \(q\sin^2 \beta+p\cos^2\beta=n\).
Divide every term by \(\cos^2 \beta\). Similar to the previous case, \(\cos^2 \beta\) cannot be zero because if it were, it would imply \(q=n\), which contradicts the given condition \(q \neq n\).
\(q \frac{\sin^2 \beta}{\cos^2 \beta} + p \frac{\cos^2 \beta}{\cos^2 \beta} = \frac{n}{\cos^2 \beta}\)
Using the same trigonometric identities (\(\tan^2 \beta\) and \(\sec^2 \beta = 1 + \tan^2 \beta\)):
\(q\tan^2 \beta + p = n\sec^2 \beta\)
Substitute \(\sec^2 \beta = 1 + \tan^2 \beta\):
\(q\tan^2 \beta + p = n(1 + \tan^2 \beta)\)
Expand the right side:
\(q\tan^2 \beta + p = n + n\tan^2 \beta\)
Rearrange the terms to isolate \(\tan^2 \beta\):
\(q\tan^2 \beta - n\tan^2 \beta = n - p\)
Factor out \(\tan^2 \beta\):
\((q - n)\tan^2 \beta = n - p\)
Since \(q \neq n\), we can divide both sides by \((q - n)\) to find \(\tan^2 \beta\):
\(\tan^2 \beta = \frac{n-p}{q-n}\)
Calculating the Ratio \(\\\frac{\tan^2 \alpha}{\tan^2 \beta}\)
Now we have the expressions for \(\tan^2 \alpha\) and \(\tan^2 \beta\). We can calculate their ratio:
\(\frac{\tan^2 \alpha}{\tan^2 \beta} = \frac{\left(\frac{m-q}{p-m}\right)}{\left(\frac{n-p}{q-n}\right)}\)
Dividing fractions involves multiplying the numerator by the reciprocal of the denominator:
\(\frac{\tan^2 \alpha}{\tan^2 \beta} = \frac{m-q}{p-m} \times \frac{q-n}{n-p}\)
To simplify this expression and match the format of the answer options, we can rewrite the terms \(p-m\) and \(q-n\). Using the property \(a-b = -(b-a)\), we have \(p-m = -(m-p)\) and \(q-n = -(n-q)\). Substituting these into the equation:
\(\frac{\tan^2 \alpha}{\tan^2 \beta} = \frac{m-q}{-(m-p)} \times \frac{-(n-q)}{n-p}\)
Now, multiply the numerators together and the denominators together:
\(\frac{\tan^2 \alpha}{\tan^2 \beta} = \frac{(m-q) \times -(n-q)}{-(m-p) \times (n-p)}\)
Notice that there is a negative sign in the numerator and a negative sign in the denominator. These two negative signs cancel each other out:
\(\frac{\tan^2 \alpha}{\tan^2 \beta} = \frac{(m-q)(n-q)}{(m-p)(n-p)}\)
Comparing with Options
The final expression we derived is \(\frac{(m-q)(n-q)}{(m-p)(n-p)}\). Let's compare this with the given options:
- Option 1: -\(\frac{(m-q)(n-q)}{(m-p)(n-p)}\)
- Option 2: -\(\frac{(m-q)(n-p)}{(m-p)(n-q)}\)
- Option 3: \(\frac{(m-q)(n-q)}{(m-p)(n-p)}\)
- Option 4: \(\frac{(m-q)(n-p)}{(m-p)(n-q)}\)
Our result perfectly matches the expression in Option 3.
Final Answer Confirmation
Based on the step-by-step derivation using trigonometric identities and algebraic manipulation, the ratio \(\frac{\tan^2 \alpha}{\tan^2 \beta}\) is equal to \(\frac{(m-q)(n-q)}{(m-p)(n-p)}\).
Similar Questions
If \(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}} = \text{cosec}\theta\), then what is \(\sin\theta\) equal to?
- If \(2 \tan\theta = \sec^2\theta-2\), where \(0 < \theta < \pi /2\), then what is \(\cot\theta\) equal to?
- If \(\alpha\) and \(\beta\) are complementary angles, then which one of the following is correct?
- What is \(\sec\theta\) equal to?
- What is \(\tan\theta\) equal to?
- What is the relation between \(p\) and \(q\)?
- What is $\tan\theta + \cot\theta$ equal to?
- Consider the following statements :
1. In a triangle ABC, if \(\sin A + \sin B + \sin C = \frac{3\sqrt{3}}{2}\), then the triangle can be equilateral.
2. In a triangle ABC, if \(\cos A + \cos B + \cos C = \frac{3}{2}\), then the triangle can be equilateral.
Which of the statements given above is/are correct? What is \((1 + \tan\alpha \tan\beta)^2 + (\tan\alpha - \tan\beta)^2\) equal to ?
Consider the following statements :
I. \(\tan 50^\circ - \cot 50^\circ\) is positive
II. \(\cot 25^\circ - \tan 25^\circ\) is negative
Which of the statements is/are correct?
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