MHT CET · Maths · Properties of Triangles
ln \(A B C(a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2}=\)
- A \(b^2\)
- B \(c^2\)
- C \(a^2 \)
- D \(a^2+b^2+c^2\)
Answer & Solution
Correct Answer
(B) \(c^2\)
Step-by-step Solution
Detailed explanation
Let \(X=(a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2} \)
\( =\left(a^2+b^2\right)\left(\cos ^2 \frac{C}{2}+\sin ^2 \frac{C}{2}\right)-2 a b\left(\cos ^2 \frac{C}{2}-\sin ^2 \frac{C}{2}\right) \)
\( =a^2+b^2-2 a b \cos C \)
\( =c^2 \quad \because(\text { cosine rule })\)
\( =\left(a^2+b^2\right)\left(\cos ^2 \frac{C}{2}+\sin ^2 \frac{C}{2}\right)-2 a b\left(\cos ^2 \frac{C}{2}-\sin ^2 \frac{C}{2}\right) \)
\( =a^2+b^2-2 a b \cos C \)
\( =c^2 \quad \because(\text { cosine rule })\)
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