AP EAMCET · Maths · Trigonometric Ratios & Identities
In a \(\triangle A B C, 2 a c \sin \frac{1}{2}(A-B+C)\) is equal to
- A \(a^2+b^2+c^2\)
- B \(a^2+b^2-c^2\)
- C \(a^2+c^2-b^2\)
- D \(b^2+c^2-a^2\)
Answer & Solution
Correct Answer
(C) \(a^2+c^2-b^2\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { In a } \triangle A B C \\ & \quad A+B+C=180^{\circ} \\ & \text { Then, } 2 a c \sin \frac{1}{2}(A-B+C) \\ & \Rightarrow 2 a c \sin \frac{1}{2}(180-B-B) \quad\left[\because A+C=180^{\circ}-B\right] \\ & \Rightarrow \quad 2 a c \sin…
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