If cosec θ - sin θ = p 3and sec θ - cos θ = q 3, then what is the value of tan θ ?
Understanding the Problem: Finding tan θ
We are given two trigonometric equations involving \(\operatorname{cosec} \theta\)< /span>, \(\sin \theta\)< /span>, \(\sec \theta\)< /span>, and \(\cos \theta\)< /span>, and we need to find the value of \(\tan \theta\)< /span> in terms of \(p\)< /span> and \(q\)< /span>. The given equations are:
- \(\operatorname{cosec} \theta - \sin \theta = p^3\)< /span>
- \(\sec \theta - \cos \theta = q^3\)< /span>
Our goal is to manipulate these equations to find an expression for \(\tan \theta\)< /span>, which is defined as \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)< /span>.
Simplifying the Equation for p
Let's start with the first equation:
\(\operatorname{cosec} \theta - \sin \theta = p^3\)< /span>
We know that \(\operatorname{cosec} \theta = \frac{1}{\sin \theta}\)< /span>. Substituting this into the equation:
\(\frac{1}{\sin \theta} - \sin \theta = p^3\)< /span>
To combine the terms on the left side, we find a common denominator:
\(\frac{1 - \sin^2 \theta}{\sin \theta} = p^3\)< /span>
Using the fundamental Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\)< /span>, we can replace \(1 - \sin^2 \theta\)< /span> with \(\cos^2 \theta\)< /span>:
\(\frac{\cos^2 \theta}{\sin \theta} = p^3\)< /span> (Equation 1')
Simplifying the Equation for q
Now, let's work with the second equation:
\(\sec \theta - \cos \theta = q^3\)< /span>
We know that \(\sec \theta = \frac{1}{\cos \theta}\)< /span>. Substituting this:
\(\frac{1}{\cos \theta} - \cos \theta = q^3\)< /span>
Finding a common denominator:
\(\frac{1 - \cos^2 \theta}{\cos \theta} = q^3\)< /span>
Again, using the identity \(\sin^2 \theta + \cos^2 \theta = 1\)< /span>, we replace \(1 - \cos^2 \theta\)< /span> with \(\sin^2 \theta\)< /span>:
\(\frac{\sin^2 \theta}{\cos \theta} = q^3\)< /span> (Equation 2')
Deriving the Value of tan θ
We now have two simplified equations:
- Equation 1': \(\frac{\cos^2 \theta}{\sin \theta} = p^3\)< /span>
- Equation 2': \(\frac{\sin^2 \theta}{\cos \theta} = q^3\)< /span>
We want to find \(\tan \theta\)< /span>, which is \(\frac{\sin \theta}{\cos \theta}\)< /span>. Let's divide Equation 2' by Equation 1' to see if we can get \(\tan \theta\)< /span>:
\(\frac{\left( \frac{\sin^2 \theta}{\cos \theta} \right)}{\left( \frac{\cos^2 \theta}{\sin \theta} \right)} = \frac{q^3}{p^3}\)< /span>
Simplify the left side by multiplying by the reciprocal of the denominator:
\(\frac{\sin^2 \theta}{\cos \theta} \times \frac{\sin \theta}{\cos^2 \theta} = \frac{q^3}{p^3}\)< /span>
Combine the terms on the left:
\(\frac{\sin^3 \theta}{\cos^3 \theta} = \frac{q^3}{p^3}\)< /span>
We know that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)< /span>, so \(\tan^3 \theta = \frac{\sin^3 \theta}{\cos^3 \theta}\)< /span>. Therefore:
\(\tan^3 \theta = \frac{q^3}{p^3}\)< /span>
To find \(\tan \theta\)< /span>, we take the cube root of both sides:
\(\tan \theta = \sqrt[3]{\frac{q^3}{p^3}}\)< /span>
\(\tan \theta = \frac{q}{p}\)< /span>
Conclusion
By simplifying the given equations for \(\operatorname{cosec} \theta - \sin \theta\)< /span> and \(\sec \theta - \cos \theta\)< /span>, and then dividing them, we successfully derived the value of \(\tan \theta\)< /span>. The value of \(\tan \theta\)< /span> is \(\frac{q}{p}\)< /span>.
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