If 2cos⁡A2=1+sin⁡A+1−sin⁡A, then A2

We have,2cos⁡A2=1+sin⁡A+1−sin⁡A⇒2cos⁡A2=cos⁡A2+sin⁡A22+cos⁡A2−sin⁡A22⇒2cos⁡A2=cos⁡A2+sin⁡A2+cos⁡A2−sin⁡A2⇒cos⁡A2+sin⁡A2≥0 and cos⁡A2−sin⁡A2≥0⇒−3π4≤A2≤3π4 and −π4≤A2≤π4⇒−π4≤A2≤π4⇒ 2nπ−π4≤A2≤2nπ+π4,n∈Z

If 2cos⁡A2=1+sin⁡A+1−sin⁡A, then A2

We have,2cos⁡A2=1+sin⁡A+1−sin⁡A⇒2cos⁡A2=cos⁡A2+sin⁡A22+cos⁡A2−sin⁡A22⇒2cos⁡A2=cos⁡A2+sin⁡A2+cos⁡A2−sin⁡A2⇒cos⁡A2+sin⁡A2≥0 and cos⁡A2−sin⁡A2≥0⇒−3π4≤A2≤3π4 and −π4≤A2≤π4⇒−π4≤A2≤π4⇒ 2nπ−π4≤A2≤2nπ+π4,n∈Z

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 $2 \cos \frac{A}{2} = \sqrt{1 + \sin A} + \sqrt{1 - \sin A},$ then $\frac{A}{2}$

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

$2 n π + \frac{π}{4}$ and $2 n π + \frac{3 π}{4}$

$2 n π - \frac{π}{4}$ and $2 n π + \frac{π}{4}$

$2 n π - \frac{3 π}{4}$ and $2 n π - \frac{π}{4}$

$- \infty and + \infty$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

We have,

$\begin{array}{l} 2 \cos \frac{A}{2} = \sqrt{1 + \sin A} + \sqrt{1 - \sin A} \\ \Rightarrow 2 \cos \frac{A}{2} = \sqrt{{(\cos \frac{A}{2} + \sin \frac{A}{2})}^{2}} + \sqrt{{(\cos \frac{A}{2} - \sin \frac{A}{2})}^{2}} \\ \Rightarrow 2 \cos \frac{A}{2} = |\cos \frac{A}{2} + \sin \frac{A}{2}| + |\cos \frac{A}{2} - \sin \frac{A}{2}| \\ \Rightarrow \cos \frac{A}{2} + \sin \frac{A}{2} \geq 0 and \cos \frac{A}{2} - \sin \frac{A}{2} \geq 0 \\ \Rightarrow - \frac{3 π}{4} \leq \frac{A}{2} \leq \frac{3 π}{4} and - \frac{π}{4} \leq \frac{A}{2} \leq \frac{π}{4} \\ \Rightarrow - \frac{π}{4} \leq \frac{A}{2} \leq \frac{π}{4} \end{array}$