MHT CET · Maths · Trigonometric Ratios & Identities
If the sides of a triangle \(a, b, c\) are in A.P., then with usual notations, a \(\cos ^2 \frac{\mathrm{C}}{2}+\mathrm{c} \cos ^2 \frac{\mathrm{~A}}{2}\) is
- A \(\frac{3 a}{2}\)
- B \(\frac{3 \mathrm{c}}{2}\)
- C \(\frac{3 b}{2}\)
- D \(\frac{\mathrm{a}+\mathrm{c}}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{3 b}{2}\)
Step-by-step Solution
Detailed explanation
Since \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in A. P.,
\(\begin{aligned}
& 2 b=a+c \\
& a \cos ^2\left(\frac{C}{2}\right)+c \cos ^2\left(\frac{A}{2}\right) \\
& =\frac{a(1+\cos C)}{2}+\frac{c(1+\cos A)}{2} \\
& =\frac{a+c+a \cos C+c \cos A}{2} \\
& =\frac{a+c+b}{2} \\
& =\frac{2 b+b}{2}=\frac{3 b}{2}
\end{aligned}\)
\(\begin{aligned}
& 2 b=a+c \\
& a \cos ^2\left(\frac{C}{2}\right)+c \cos ^2\left(\frac{A}{2}\right) \\
& =\frac{a(1+\cos C)}{2}+\frac{c(1+\cos A)}{2} \\
& =\frac{a+c+a \cos C+c \cos A}{2} \\
& =\frac{a+c+b}{2} \\
& =\frac{2 b+b}{2}=\frac{3 b}{2}
\end{aligned}\)
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