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MHT CET · Maths · Trigonometric Equations

In a triangle ABC , with usual notations, \(\frac{\cos B+\cos C}{b+c}+\frac{\cos A}{a}\) has the value

  1. A \(\frac{1}{\mathrm{~b}+\mathrm{c}}\)
  2. B \(\cdot \frac{1}{\mathrm{~b}}\)
  3. C \(\frac{1}{\mathrm{c}}\)
  4. D \(\frac{1}{\mathrm{a}}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{1}{\mathrm{a}}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \frac{\cos B+\cos C}{b+c}+\frac{\cos A}{a} \\ & =\frac{a \cos B+a \cos C+b \cos A+c \cos A}{a(b+c)} \\ & =\frac{(a \cos B+b \cos A)+(a \cos C+c \cos A)}{a(b+c)}\end{aligned}\)
\(\begin{aligned}
& =\frac{c+b}{a(b+c)} \\
& =\frac{1}{a}
\end{aligned}\)
...[By projection rule]