Question

Prove that : \(\tan^2 \theta + \cot^2 \theta + 2 = \sec^2 \theta \csc^2 \theta\)

Hint:

An alternative way to prove this is using standard identity replacements:
- Replace \(\tan^2 \theta\) with \(\sec^2 \theta - 1\)
- Replace \(\cot^2 \theta\) with \(\csc^2 \theta - 1\)
\[ \text{LHS} = (\sec^2 \theta - 1) + (\csc^2 \theta - 1) + 2 = \sec^2 \theta + \csc^2 \theta \]
Convert to sines and cosines:
\[ \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta} = \sec^2 \theta \csc^2 \theta \]
This is also a highly elegant and quick route to the proof!

Answer 1

Correct Answer: 4

Step 1: Understanding the Question:
We are given a trigonometric identity:
\[ \tan^2 \theta + \cot^2 \theta + 2 = \sec^2 \theta \csc^2 \theta \]
We need to prove that the Left-Hand Side (LHS) is equal to the Right-Hand Side (RHS).

Step 2: Key Formula or Approach:
We can approach this proof using standard algebraic identities and trigonometric relations:
1. We know that \(\tan \theta \cdot \cot \theta = 1\).
2. Therefore, the LHS can be rewritten as a perfect square:
\[ \tan^2 \theta + \cot^2 \theta + 2(\tan \theta \cot \theta) = (\tan \theta + \cot \theta)^2 \]
3. Express \(\tan \theta\) and \(\cot \theta\) in terms of \(\sin \theta\) and \(\cos \theta\) to simplify.
4. Use the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\).

Step 3: Detailed Explanation:

• Let us write the Left-Hand Side (LHS) of the identity:
\[ \text{LHS} = \tan^2 \theta + \cot^2 \theta + 2 \]

• Since \(\tan \theta \cdot \cot \theta = 1\), we can write the constant term 2 as \(2(\tan \theta \cot \theta)\):
\[ \text{LHS} = \tan^2 \theta + \cot^2 \theta + 2\tan \theta \cot \theta \]

• This matches the perfect square expansion \(a^2 + b^2 + 2ab = (a+b)^2\):
\[ \text{LHS} = (\tan \theta + \cot \theta)^2 \]

• Now, let us simplify the expression inside the parentheses by converting to sines and cosines:
\[ \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \]

• Take a common denominator to add the fractions:
\[ \tan \theta + \cot \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \]

• Substitute the identity \(\sin^2 \theta + \cos^2 \theta = 1\) in the numerator:
\[ \tan \theta + \cot \theta = \frac{1}{\sin \theta \cos \theta} \]

• Convert the reciprocal terms into secant and cosecant functions:
\[ \tan \theta + \cot \theta = \sec \theta \csc \theta \]

• Substitute this back into our squared expression for LHS:
\[ \text{LHS} = (\tan \theta + \cot \theta)^2 \] \[ \text{LHS} = (\sec \theta \csc \theta)^2 \] \[ \text{LHS} = \sec^2 \theta \csc^2 \theta \]

• Compare with the Right-Hand Side (RHS):
\[ \text{RHS} = \sec^2 \theta \csc^2 \theta \] - Since \(\text{LHS} = \text{RHS}\), the identity is proven.


Step 4: Final Answer:
The identity is proven successfully.