Get the answer to "Prove that tan−1(xsin⁡α1−xcos⁡α)−tan−1(xcos⁡αsin⁡α) is independent of x and..."

cindakayumn

Answered question

2022-02-27

Prove that

\tan^{- 1} (\frac{x \sin α}{1 - x \cos α}) - \tan^{- 1} (\frac{x \cos α}{\sin α})

is independent of x and is equal to

\frac{π}{2} - α

tardanetkd2

Beginner2022-02-28Added 9 answers

The '-' sign is missing

\tan^{- 1} (\frac{x \sin α}{1 - x \cos α}) - \tan^{- 1} (\frac{x - \cos α}{\sin α})

= \tan^{- 1} (\frac{\frac{x \sin α}{1 - x \cos α} - \frac{x \cos α}{\sin α}}{1 + \frac{x \sin α}{1 - x \cos α} \cdot \frac{x - \cos α}{\sin α}})

= \tan^{- 1} (\frac{x \sin α (\sin α) - (x - \cos α) (1 - x \cos a l p h)}{(1 - x \cos α) \sin α + x \sin α (x - \cos α)})

assuming

(1 - x \cos α) \sin α \neq 0

= \tan^{- 1} (\frac{x (\sin^{2} α - 1 - \cos^{2} α) + x^{2} \cos α + \cos α}{\sin α (1 - 2 x \cos α + x^{2})})

= \tan^{- 1} (\frac{\cos α (1 - 2 x \cos α + x^{2})}{\sin (α (1 - 2 x \cos α + x^{2})}}

using

\sin^{2} α - 1 = - \cos^{2} α

= \tan^{- 1} (\cot α)

assuming

1 - 2 x \cos α + x^{2} \neq 0

= \tan^{- 1} (\tan (\frac{π}{2} - α)) = \frac{π}{2} - α

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