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Mathematics Question on Trigonometric Identities

cos⁡ xcos⁡ x−2y=λ ⇒ tan⁡ x−y\frac{\cos \, x}{\cos \, x -2y} = \lambda \, \Rightarrow \, \tan \, x - y is equal to

A

1+λ1−λ\frac{1 + \lambda}{1 - \lambda}

B

1−λ1+λ\frac{1 - \lambda}{1 + \lambda}

C

λ1+λ\frac{ \lambda}{1 + \lambda}

D

λ1−λ\frac{ \lambda}{1 - \lambda}

Answer

1−λ1+λ\frac{1 - \lambda}{1 + \lambda}

Explanation

Solution

Now, tan⁡(x−y) tan⁡ y\tan (x - y) \, \tan \, y
=sin⁡ x−y sin⁡ ycos⁡ x−y cos⁡ y×22= \frac{\sin \, x-y \, \sin \, y}{\cos \, x-y \, \cos \, y} \times \frac{2}{2}
=cos⁡ x−2y−cos⁡ xcos⁡ x−2y+cos⁡ x= \frac{\cos \, x-2y - \cos \, x}{\cos \, x - 2y + \cos \, x}
=1−cos⁡ xcos⁡ x−2y1+cos⁡ xcos⁡ x−2y= \frac{1 - \frac{\cos \, x}{\cos \, x - 2y}}{ 1 + \frac{\cos \, x}{ \cos \, x -2y}}
=1−λ1+λ= \frac{1 - \lambda}{ 1 +\lambda}
Given , λ=cos⁡ xcos⁡ x−2y\lambda = \frac{\cos \, x}{ \cos \, x - 2y}