Question:
Let \(f(x)=\frac{sinx+cosx-\sqrt2}{sinx-cosx},x∈[0.\pi]-\left[\frac{\pi}{4}\right]. Then f\left(\frac{7\pi}{12}\right)f^n\left(\frac{7\pi}{12}\right)\)is equal to
Let \(f(x)=\frac{sinx+cosx-\sqrt2}{sinx-cosx},x∈[0.\pi]-\left[\frac{\pi}{4}\right]. Then f\left(\frac{7\pi}{12}\right)f^n\left(\frac{7\pi}{12}\right)\)is equal to
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Use trigonometric identities to simplify expressions before substituting values for easier calculations.
Updated On: Mar 25, 2026
- \(\frac{-2}{3}\)
- \(\frac{-1}{3\sqrt3}\)
- \(\frac{2}{3\sqrt3}\)
- \(\frac{2}{9}\)
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The Correct Option is D
Solution and Explanation
We are given the function:
\[
f(x) = - \tan\left( \frac{x}{2} - \frac{\pi}{8} \right)
\]
Now, differentiate the function \( f(x) \):
\[
f'(x) = - \frac{1}{2} \sec^2\left( \frac{x}{2} - \frac{\pi}{8} \right)
\]
Now, calculate the second derivative:
\[
f''(x) = - \sec^2\left( \frac{x}{2} - \frac{\pi}{8} \right) \cdot \tan\left( \frac{x}{2} - \frac{\pi}{8} \right) \cdot \frac{1}{2}
\]
Now, calculate the value of \( f\left( \frac{7\pi}{12} \right) \):
\[
f\left( \frac{7\pi}{12} \right) = - \tan\left( \frac{\pi}{6} \right) = - \frac{1}{\sqrt{3}}.
\]
Next, calculate the second derivative at \( x = \frac{7\pi}{12} \):
\[
f''\left( \frac{7\pi}{12} \right) = - \frac{1}{2} \sec^2\left( \frac{\pi}{6} \right) \cdot \tan\left( \frac{\pi}{6} \right) = - \frac{1}{2} \times \frac{4}{3} \times \frac{1}{\sqrt{3}} = \frac{-4}{3\sqrt{3}}.
\]
Finally, we calculate the product:
\[
f\left( \frac{7\pi}{12} \right) \cdot f''\left( \frac{7\pi}{12} \right) = \left( -\frac{1}{\sqrt{3}} \right) \cdot \left( \frac{-4}{3\sqrt{3}} \right) = \frac{2}{3}.
\]
Final Answer: \( \frac{2}{3} \).
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