JEE Mains · Maths · STD 11 - Trigonometrical equations
In a triangle \(A B C\), if \(\cos A+2 \cos B+\cos C=2\) and the lengths of the sides opposite to the angles \(A\) and \(C\) are \(3\) and \(7\) respectively, then \(\cos A-\cos\) \(C\) is equal to
- A \(\frac{3}{7}\)
- B \(\frac{9}{7}\)
- C \(\frac{10}{7}\)
- D \(\frac{5}{7}\)
Answer & Solution
Correct Answer
(C) \(\frac{10}{7}\)
Step-by-step Solution
Detailed explanation
\(\cos A+\cos C=2(1-\cos B)\) \(2 \cos \frac{A+C}{2} \cos \frac{A-C}{2}=4 \sin ^2 B / 2\) \(\text { as } \cos \left(\frac{A+C}{2}\right)=\sin \frac{B}{2}\) \(\text { so } \cos \frac{ A - C }{2}=2 \sin \frac{ B }{2}\)…
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