First slide

Trigonometric transformations

Question

If A,B,C are the angles of a triangle and $c os B + \cos C = 4 \sin^{2} \frac{A}{2}$ , then $\tan \frac{B}{2} \tan \frac{C}{2}$ is equal to

Moderate

A

1/2
B

1/3
C

2/3
D

3/2

Solution

$\cos B + \cos C = 4 \sin^{2} (A / 2)$
$\begin{array}{l} \Rightarrow 2 \cos \frac{B + C}{2} \cos \frac{B - C}{2} = 4 \sin \frac{A}{2} \sin (\frac{π}{2} - \frac{B + C}{2}) \\ \Rightarrow 2 \sin \frac{A}{2} \cos \frac{B - C}{2} = 4 \sin \frac{A}{2} \cos \frac{B + C}{2} \\ \Rightarrow \cos \frac{B}{2} \cos \frac{C}{2} + \sin \frac{B}{2} \sin \frac{C}{2} = \end{array}$
$\begin{array}{l} 2 (\cos \frac{B}{2} \cos \frac{C}{2} - \sin \frac{B}{2} \sin \frac{C}{2}) \\ \Rightarrow 3 \sin \frac{B}{2} \sin \frac{C}{2} = \cos \frac{B}{2} \cos \frac{C}{2} \\ \Rightarrow \tan \frac{B}{2} \tan \frac{C}{2} = \frac{1}{3} . \end{array}$

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