cot−1⁡(cos⁡α)−tan−1⁡(cos⁡α)=x ⇒π2−2tan−1⁡(cos⁡α)=x⇒sin⁡x=sin⁡π2−2tan−1⁡(cos⁡α)put tan−1⁡(cos⁡α)=θ⇒tan⁡θ=cos⁡αsin⁡x=sin⁡π2−2θ=cos⁡2θ=1−tan2⁡θ1+tan2⁡θ=1−cos⁡α1+cos⁡αsin⁡x=1−cos⁡α1+cos ⁡α=2sin2⁡α/22cos2⁡α/2⇒sin⁡x=tan2⁡α2

Download Pdf

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

$\cot^{- 1} (\sqrt{\cos α}) - {Tan}^{- 1} (\sqrt{\cos α}) = x \geq 0, then \sin x =$

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

$\tan^{2} \frac{α}{2}$

$\cot \frac{α}{2}$

$\tan α$

$\cot^{2} (\frac{α}{2})$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \cot^{- 1} (\sqrt{\cos α}) - \tan^{- 1} (\sqrt{\cos α}) = x \\ \Rightarrow \frac{π}{2} - 2 \tan^{- 1} (\sqrt{\cos α}) = x \\ \Rightarrow \sin x = \sin (\frac{π}{2} - 2 \tan^{- 1} (\sqrt{\cos α})) \end{array}$
$put \tan^{- 1} (\sqrt{\cos α}) = θ$
$\begin{array}{l} \Rightarrow \tan θ = \sqrt{\cos α} \\ \sin x = \sin (\frac{π}{2} - 2 θ) = \cos 2 θ \\ = \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ} \\ = \frac{1 - \cos α}{1 + \cos α} \\ \sin x = \frac{1 - \cos α}{1 + \cos^{} α} \\ = \frac{2 \sin^{2} α / 2}{2 \cos^{2} α / 2} \\ \Rightarrow \sin x = \tan^{2} \frac{α}{2} \end{array}$