If tanx2=cosecx-sinx,then sec2x2=

Let t=tanx2, ∴ t=1sinx-sinx=1+t22t-2t1+t2=(1-t2)22t(1+t2) ⇒ 2t2(1+t2)=(1-t2)2⇒ t4+4t2-1=0 Solving, t2=5-2 , Now sec2x2=t2+1=5-1

If tanx2=cosecx-sinx,then sec2x2=

Let t=tanx2, ∴ t=1sinx-sinx=1+t22t-2t1+t2=(1-t2)22t(1+t2) ⇒ 2t2(1+t2)=(1-t2)2⇒ t4+4t2-1=0 Solving, t2=5-2 , Now sec2x2=t2+1=5-1

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

If $\tan \frac{x}{2} = \cos e c x - \sin x,$ then $s e c^{2} \frac{x}{2} =$

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

$\sqrt{5}$ + 1

$\sqrt{5}$ - 1

$\sqrt{5}$ - 2

$\sqrt{5}$ + 2

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$L e t t = \tan \frac{x}{2}, ∴ t = \frac{1}{\sin x} - \sin x = \frac{1 + t^{2}}{2 t} - \frac{2 t}{1 + t^{2}} = \frac{{(1 - t^{2})}^{2}}{2 t (1 + t^{2})} \Rightarrow 2 t^{2} (1 + t^{2}) = {(1 - t^{2})}^{2} \Rightarrow t^{4} + 4 t^{2} - 1 = 0 S o l v i n g, t^{2} = \sqrt{5} - 2, N o w s e c^{2} \frac{x}{2} = t^{2} + 1 = \sqrt{5} - 1$