MHT CET · Maths · Trigonometric Ratios & Identities
In a triangle \(\mathrm{ABC}\), with usual notations, if \(\mathrm{c}=4\), then value of \((a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2}\) is
- A \(4\)
- B \(16\)
- C \(9\)
- D \(2\)
Answer & Solution
Correct Answer
(B) \(16\)
Step-by-step Solution
Detailed explanation
\((a-b)^2 \cos ^2 \frac{C}{2}+(a+b)^2 \sin ^2 \frac{C}{2} \)
\( =\left(a^2-2 a b+b^2\right) \cos ^2 \frac{C}{2}+\left(a^2+2 a b+b^2\right) \sin ^2 \frac{C}{2} \)
\( =\left(a^2+b^2\right)\left(\cos ^2 \frac{C}{2}+\sin ^2 \frac{C}{2}\right) \)
\( -2 a b \cos ^2 \frac{C}{2}+2 a b \sin ^2 \frac{C}{2} \)
\( =a^2+b^2-2 a b\left(\cos ^2 \frac{C}{2}-\sin ^2 \frac{C}{2}\right) \)
\( =a^2+b^2-2 a b \cdot \cos C \)
\( =a^2+b^2-\left(a^2+b^2-c^2\right) \ldots[\text { By Cosine rule }] \)
\( =c^2=4^2=16\)
\( =\left(a^2-2 a b+b^2\right) \cos ^2 \frac{C}{2}+\left(a^2+2 a b+b^2\right) \sin ^2 \frac{C}{2} \)
\( =\left(a^2+b^2\right)\left(\cos ^2 \frac{C}{2}+\sin ^2 \frac{C}{2}\right) \)
\( -2 a b \cos ^2 \frac{C}{2}+2 a b \sin ^2 \frac{C}{2} \)
\( =a^2+b^2-2 a b\left(\cos ^2 \frac{C}{2}-\sin ^2 \frac{C}{2}\right) \)
\( =a^2+b^2-2 a b \cdot \cos C \)
\( =a^2+b^2-\left(a^2+b^2-c^2\right) \ldots[\text { By Cosine rule }] \)
\( =c^2=4^2=16\)
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