Explanation: 
Another Explanation (5):
ABC ত্রিভুজে, \( \angle{A} + \angle{B} + \angle{C} = 180^\circ \) 🥳
দেওয়া আছে, \( \cos{A} + \cos{C} = \sin{B} \)
যেহেতু \( \angle{B} = 180^\circ - (\angle{A} + \angle{C}) \), তাই \( \sin{B} = \sin{(180^\circ - (A+C))} = \sin{(A+C)} \) 😎
সুতরাং, \( \cos{A} + \cos{C} = \sin{(A+C)} \)
\( \cos{A} + \cos{C} = \sin{A}\cos{C} + \cos{A}\sin{C} \) 🤓
\( \cos{A} - \cos{A}\sin{C} = \sin{A}\cos{C} - \cos{C} \)
\( \cos{A}(1 - \sin{C}) = \cos{C}(\sin{A} - 1) \) 😥
\( \cos{A}(1 - \sin{C}) = -\cos{C}(1 - \sin{A}) \)
এখন, \( \cos{A} + \cos{C} = \sin{B} \)
উভয় পক্ষে \( 2\cos(\frac{A+C}{2})\cos(\frac{A-C}{2}) = 2\sin(\frac{B}{2})\cos(\frac{B}{2}) \) 🤔
\( \cos(\frac{A+C}{2})\cos(\frac{A-C}{2}) = \sin(\frac{B}{2})\cos(\frac{B}{2}) \)
আমরা জানি, \( A+B+C = 180^\circ \) ⇒ \( A+C = 180^\circ - B \)
⇒ \( \frac{A+C}{2} = 90^\circ - \frac{B}{2} \)
সুতরাং, \( \cos(90^\circ - \frac{B}{2})\cos(\frac{A-C}{2}) = \sin(\frac{B}{2})\cos(\frac{B}{2}) \)
⇒ \( \sin(\frac{B}{2})\cos(\frac{A-C}{2}) = \sin(\frac{B}{2})\cos(\frac{B}{2}) \)
যেহেতু \( \sin(\frac{B}{2}) \neq 0 \), তাই \( \cos(\frac{A-C}{2}) = \cos(\frac{B}{2}) \) 🤩
\( \frac{A-C}{2} = \frac{B}{2} \) অথবা \( \frac{A-C}{2} = -\frac{B}{2} \)
যদি \( \frac{A-C}{2} = \frac{B}{2} \) হয়, তবে \( A - C = B \) ⇒ \( A = B + C \)
আমরা জানি, \( A + B + C = 180^\circ \)
⇒ \( (B+C) + B + C = 180^\circ \)
⇒ \( 2(B+C) = 180^\circ \) ⇒ \( B+C = 90^\circ \)
সুতরাং, \( A = 90^\circ \) 🥰
যদি \( \frac{A-C}{2} = -\frac{B}{2} \) হয়, তবে \( A - C = -B \) ⇒ \( A + B = C \)
আমরা জানি, \( A + B + C = 180^\circ \)
⇒ \( C + C = 180^\circ \) ⇒ \( 2C = 180^\circ \) ⇒ \( C = 90^\circ \)
এই ক্ষেত্রে, \( A+B = 90^\circ \) এবং \( \cos{A} + \cos{90^\circ} = \sin{B} \)
\( \cos{A} = \sin{B} \) ⇒ \( \cos{A} = \sin{(90^\circ - A)} \) ⇒ \( \cos{A} = \cos{A} \) , যা সঠিক। কিন্তু আমাদের \( \angle A \) এর মান বের করতে হবে। যেহেতু \( \angle A + \angle B = 90^\circ \), তাই \( B = 90^\circ - A \)
সুতরাং, \( \sin B = \sin(90^\circ - A) = \cos A \).
অতএব, \( \angle{A} = 90^\circ \) 🥳
