Institution: বঙ্গবন্ধু শেখ মুজিবুর রহমান বিজ্ঞান ও প্রযুক্তি বিশ্ববিদ্যালয় (BSMRSTU) Exam: A ইউনিট || বঙ্গবন্ধু শেখ মুজিবুর রহমান বিজ্ঞান ও প্রযুক্তি বিশ্ববিদ্যালয় || 2011
Topic: Mathematics (সাধারণ গণিত), Subject: Mathematics (সাধারণ গণিত)

Best Answer Explained in Below-:

To evaluate the expression \( \sin 65^\circ + \cos 65^\circ \), we can use a trigonometric identity involving the sine of sums. We start by observing that \( \sin 65^\circ \) can be rewritten using the identity for sine:

\[
\sin 65^\circ = \sin(45^\circ + 20^\circ) = \sin 45^\circ \cos 20^\circ + \cos 45^\circ \sin 20^\circ = \frac{\sqrt{2}}{2} \cos 20^\circ + \frac{\sqrt{2}}{2} \sin 20^\circ = \frac{\sqrt{2}}{2} (\cos 20^\circ + \sin 20^\circ)
\]

Similarly, for \( \cos 65^\circ \):

\[
\cos 65^\circ = \cos(45^\circ + 20^\circ) = \cos 45^\circ \cos 20^\circ - \sin 45^\circ \sin 20^\circ = \frac{\sqrt{2}}{2} \cos 20^\circ - \frac{\sqrt{2}}{2} \sin 20^\circ = \frac{\sqrt{2}}{2} (\cos 20^\circ - \sin 20^\circ)
\]

Adding these two expressions together gives:

\[
\sin 65^\circ + \cos 65^\circ = \frac{\sqrt{2}}{2} (\cos 20^\circ + \sin 20^\circ) + \frac{\sqrt{2}}{2} (\cos 20^\circ - \sin 20^\circ)
\]

This simplifies to:

\[
\sin 65^\circ + \cos 65^\circ = \sqrt{2} \cos 20^\circ
\]

However, since we are combining the components with the identities, we realize we're also able to relate this by direct cosine products. Thus, the expression reduces to \( \sqrt{2} \cos 20^\circ \), confirming that the correct answer is indeed **B) \( \sqrt{2} \cos 20^\circ \)**.