The value of tan−1⁡xcos⁡θ1−xsin⁡θ−cot−1⁡cos⁡θx−sin⁡θ is

tan−1⁡xcos⁡θ1−xsin⁡θ−cot−1⁡cos⁡θx−sin⁡θ=tan−1⁡xcos⁡θ1−xsin⁡θ−tan−1⁡x−sin⁡θcos⁡θ=tan−1⁡xcos⁡θ1−xsin⁡θ−x−sin⁡θcos⁡θ1+xcos⁡θ1−xsin⁡θx−sin⁡θcos⁡θ=tan−1⁡xcos2⁡θ−x+sin⁡θ+x2sin⁡θ−xsin2⁡θcos⁡θ−xcos⁡θsin⁡θ+x2cos⁡θ−xcos⁡θsin⁡θ=tan−1⁡−xsin2⁡θ+sin⁡θ+x2sin⁡θ−xsin2⁡θcos⁡θ−2xcos⁡θsin⁡θ+x2cos⁡θ=tan−1⁡−2xsin2⁡θ+sin⁡θ+x2sin⁡θcos⁡θ−2xcos⁡θsin⁡θ+x2cos⁡θ=tan−1⁡sin⁡θ−2xsin⁡θ+1+x2cos⁡θ1−2xsin⁡θ+x2=tan−1⁡(tan⁡θ)=θ

The value of tan−1⁡xcos⁡θ1−xsin⁡θ−cot−1⁡cos⁡θx−sin⁡θ is

tan−1⁡xcos⁡θ1−xsin⁡θ−cot−1⁡cos⁡θx−sin⁡θ=tan−1⁡xcos⁡θ1−xsin⁡θ−tan−1⁡x−sin⁡θcos⁡θ=tan−1⁡xcos⁡θ1−xsin⁡θ−x−sin⁡θcos⁡θ1+xcos⁡θ1−xsin⁡θx−sin⁡θcos⁡θ=tan−1⁡xcos2⁡θ−x+sin⁡θ+x2sin⁡θ−xsin2⁡θcos⁡θ−xcos⁡θsin⁡θ+x2cos⁡θ−xcos⁡θsin⁡θ=tan−1⁡−xsin2⁡θ+sin⁡θ+x2sin⁡θ−xsin2⁡θcos⁡θ−2xcos⁡θsin⁡θ+x2cos⁡θ=tan−1⁡−2xsin2⁡θ+sin⁡θ+x2sin⁡θcos⁡θ−2xcos⁡θsin⁡θ+x2cos⁡θ=tan−1⁡sin⁡θ−2xsin⁡θ+1+x2cos⁡θ1−2xsin⁡θ+x2=tan−1⁡(tan⁡θ)=θ

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The value of $\tan^{- 1} (\frac{xcos θ}{1 - xsin θ}) - \cot^{- 1} (\frac{\cos θ}{x - \sin θ})$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

2 $θ$

$θ$

$\frac{θ}{2}$

independent of $θ$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \tan^{- 1} (\frac{xcos θ}{1 - xsin θ}) - \cot^{- 1} (\frac{\cos θ}{x - \sin θ}) \\ = \tan^{- 1} (\frac{xcos θ}{1 - xsin θ}) - \tan^{- 1} (\frac{x - \sin θ}{\cos θ}) \\ = \tan^{- 1} [\frac{\frac{xcos θ}{1 - xsin θ} - \frac{x - \sin θ}{\cos θ}}{1 + (\frac{xcos θ}{1 - xsin θ}) (\frac{x - \sin θ}{\cos θ})}] \\ = \tan^{- 1} (\frac{{xcos}^{2} θ - x + \sin θ + x^{2} \sin θ - {xsin}^{2} θ}{\cos θ - xcos θsin θ + x^{2} \cos θ - xcos θsin θ}) \\ = \tan^{- 1} (\frac{- {xsin}^{2} θ + \sin θ + x^{2} \sin θ - {xsin}^{2} θ}{\cos θ - 2 xcos θsin θ + x^{2} \cos θ}) \\ = \tan^{- 1} (\frac{- 2 {xsin}^{2} θ + \sin θ + x^{2} \sin θ}{\cos θ - 2 xcos θsin θ + x^{2} \cos θ}) \\ = \tan^{- 1} \{\frac{\sin θ (- 2 xsin θ + 1 + x^{2})}{\cos θ (1 - 2 xsin θ + x^{2})}\} \\ = \tan^{- 1} (\tan θ) = θ \end{array}$