Aspire Faculty ID #11640 · Topic: NIMCET 2024 · Just now
NIMCET 2024

The value of $\tan \Bigg{(}\frac{\pi}{4}+\theta\Bigg{)}\tan \Bigg{(}\frac{3\pi}{4}+\theta\Bigg{)}$ is

Solution

We are given:

\[ \text{Evaluate } \tan\left(\frac{\pi}{4} + \theta\right) \cdot \tan\left(\frac{3\pi}{4} + \theta\right) \]

✳ Step 1: Use identity

\[ \tan\left(A + B\right) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \] But we don’t need expansion — use known angle values:

\[ \tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan\theta}{1 - \tan\theta} \]

\[ \tan\left(\frac{3\pi}{4} + \theta\right) = \frac{-1 + \tan\theta}{1 + \tan\theta} \]

✳ Step 2: Multiply

\[ \left(\frac{1 + \tan\theta}{1 - \tan\theta}\right) \cdot \left(\frac{-1 + \tan\theta}{1 + \tan\theta}\right) \]

Simplify:

\[ = \frac{(1 + \tan\theta)(-1 + \tan\theta)}{(1 - \tan\theta)(1 + \tan\theta)} = \frac{(\tan^2\theta - 1)}{1 - \tan^2\theta} = \boxed{-1} \]

✅ Final Answer:

\[ \boxed{-1} \]

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