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Q.

If $\cos θ_{1} = 2 \cos θ_{2},$ then $\tan \frac{θ_{1} - θ_{2}}{2}$ . $\tan \frac{θ_{1} + θ_{2}}{2}$ is equal to

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a

\frac{1}{3}

b

\frac{- 1}{3}

c

1

d

- 1

answer is B.

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Detailed Solution

It is given that,

\cos θ_{1} = 2 \cos θ_{2}

.
Some expressions can be deduced as follows,

\sin \frac{A - B}{2} \times \sin \frac{A + B}{2} = \sin^{2} \frac{A}{2} - \sin^{2} \frac{B}{2} = 1 - cosA - 1 + cosB = cosB - cosA

\cos \frac{A - B}{2} \times \cos \frac{A + B}{2} = \cos^{2} \frac{B}{2} - \sin^{2} \frac{A}{2} = 1 + cosB - 1 + cosA = cosB + cosA

To find the value of the expression

\tan \frac{θ_{1} - θ_{2}}{2} \tan \frac{θ_{1} + θ_{2}}{2}

, we can simplify it as follows

= \frac{\sin \frac{θ_{1} - θ_{2}}{2} \sin \frac{θ_{1} + θ_{2}}{2}}{\cos \frac{θ_{1} - θ_{2}}{2} \cos \frac{θ_{1} + θ_{2}}{2}}

= \frac{\cos θ_{2} - \cos θ_{1}}{\cos θ_{2} + \cos θ_{1}}

= \frac{\cos θ_{2} - 2 \cos θ_{2}}{\cos θ_{2} + 2 \cos θ_{2}}

= \frac{- \cos θ_{2}}{3 \cos θ_{2}}

= \frac{- 1}{3}

So, the value of

\tan \frac{θ_{1} - θ_{2}}{2}

.

\tan \frac{θ_{1} + θ_{2}}{2} = \frac{- 1}{3}

.
Hence, the correct option is (2).

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