মান নির্ণয় কর: tan(17π/4)cos(-11π/4) + sec(-34π/3)cosec(25π/6)

প্রদত্ত অভিব্যক্তিটি হল:

\[
\text{tan}\left(\frac{17\pi}{4}\right) \cos\left(-\frac{11\pi}{4}\right) + \sec\left(-\frac{34\pi}{3}\right) \csc\left(\frac{25\pi}{6}\right)
\]

প্রথমে, প্রতিটি কোণকে \(2\pi\) বা \(360^\circ\) এর গুণিতকের মধ্যে রূপান্তর করি:

1. **\( \tan\left(\frac{17\pi}{4}\right) \)**:
   \[
   \frac{17\pi}{4} = 4\pi + \frac{\pi}{4} = 2(2\pi) + \frac{\pi}{4}
   \]
   যেহেতু \(2\pi\) এর গুণিতক যোগ করলে কোণের মান অপরিবর্তিত থাকে, তাই:
   \[
   \tan\left(\frac{17\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1
   \]

2. **\( \cos\left(-\frac{11\pi}{4}\right) \)**:
   \[
   -\frac{11\pi}{4} = -2\pi - \frac{3\pi}{4} = -\frac{3\pi}{4} \quad (\text{যেহেতু } -2\pi \text{ যোগ করলে কোণের মান অপরিবর্তিত থাকে})
   \]
   \[
   \cos\left(-\frac{3\pi}{4}\right) = \cos\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}}
   \]

3. **\( \sec\left(-\frac{34\pi}{3}\right) \)**:
   \[
   -\frac{34\pi}{3} = -\frac{36\pi}{3} + \frac{2\pi}{3} = -12\pi + \frac{2\pi}{3}
   \]
   যেহেতু \( -12\pi \) হল \( 2\pi \) এর গুণিতক, তাই:
   \[
   \sec\left(-\frac{34\pi}{3}\right) = \sec\left(\frac{2\pi}{3}\right) = -\sec\left(\frac{\pi}{3}\right) = -2
   \]

4. **\( \csc\left(\frac{25\pi}{6}\right) \)**:
   \[
   \frac{25\pi}{6} = 4\pi + \frac{\pi}{6} = 2(2\pi) + \frac{\pi}{6}
   \]
   \[
   \csc\left(\frac{25\pi}{6}\right) = \csc\left(\frac{\pi}{6}\right) = 2
   \]

এখন, সমস্ত উপাদানকে একত্রিত করি:

\[
\tan\left(\frac{17\pi}{4}\right) \cos\left(-\frac{11\pi}{4}\right) + \sec\left(-\frac{34\pi}{3}\right) \csc\left(\frac{25\pi}{6}\right) = 1 \times \left(-\frac{1}{\sqrt{2}}\right) + (-2) \times 2
\]

\[
= -\frac{1}{\sqrt{2}} - 4
\]

\[
= -\frac{1}{\sqrt{2}} - \frac{4\sqrt{2}}{\sqrt{2}} = -\frac{1 + 4\sqrt{2}}{\sqrt{2}}
\]

তাহলে চূড়ান্ত মান:

\[
= -\frac{1 + 4\sqrt{2}}{\sqrt{2}}
\]

আপনার আরও বিস্তারিত জানা বা কোনো প্রশ্ন আছে কি?

**Related Questions**:
1. What is the value of \(\tan\left(\frac{\pi}{4}\right)\) and why?
2. How do you simplify trigonometric functions with negative angles?
3. What is the periodicity of the secant function?
4. How do you convert any angle to the first quadrant equivalent?
5. Why does \( \csc(\frac{\pi}{6}) \) equal 2?

**Tip**: Remember that the period of the tangent function is \( \pi \) and for cosine and sine, it's \( 2\pi \).

Learn how to evaluate the expression tan(17π/4)cos(-11π/4) + sec(-34π/3)cosec(25π/6) step-by-step with detailed explanations and trigonometric conversions. This problem involves advanced trigonometric functions and their evaluations. Suitable for advanced high school students and beyond.

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