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: Start with the Left-Hand Side (LHS).
We need to prove: $\tan^2\theta + \cot^2\theta + 2 = \sec^2\theta \cdot \csc^2\theta$. Let us work on the LHS: \[ \text{LHS} = \tan^2\theta + \cot^2\theta + 2 \]
Step 2: Express $\tan\theta$ and $\cot\theta$ in Terms of $\sin$ and $\cos$.
\[ \tan\theta = \frac{\sin\theta}{\cos\theta}, \quad \cot\theta = \frac{\cos\theta}{\sin\theta} \] So: \[ \text{LHS} = \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} + 2 \]
Step 3: Combine Into a Single Fraction.
The common denominator is $\sin^2\theta \cos^2\theta$: \[ \text{LHS} = \frac{\sin^4\theta + \cos^4\theta + 2\sin^2\theta\cos^2\theta}{\sin^2\theta\cos^2\theta} \]
Step 4: Recognise the Perfect Square in the Numerator.
\[ \sin^4\theta + \cos^4\theta + 2\sin^2\theta\cos^2\theta = (\sin^2\theta + \cos^2\theta)^2 = 1^2 = 1 \] This is because $(a^2 + b^2 + 2ab) = (a+b)^2$ with $a = \sin^2\theta$ and $b = \cos^2\theta$.
Step 5: Simplify the LHS.
\[ \text{LHS} = \frac{1}{\sin^2\theta \cos^2\theta} = \frac{1}{\sin^2\theta} \times \frac{1}{\cos^2\theta} = \csc^2\theta \cdot \sec^2\theta \]
Step 6: Conclude the Proof.
We have shown LHS $= \sec^2\theta \cdot \csc^2\theta =$ RHS. Hence proved. \[ \boxed{\tan^2\theta + \cot^2\theta + 2 = \sec^2\theta \cdot \csc^2\theta} \]