TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

(sin−1y)2+cos−1x=‌

nπ2

4

and (cos−1x)(sin−1,y)2=‌

π4

16

Let cos−1x=a and sin−1y=b⇒a+b2=‌

nπ2

4

.... (i)
and a⋅b2=‌

π4

16

...(ii)
Now, least value of n can be 2
Adding Eqs. (i) and (iii), we get
2a=‌

nπ2

4

or 2a=‌

2π2

4

‌‌(∵n=2)
or Now, from Eq. (iii), a=b
(sin−1y)2=‌

π2

4

or sin−1y=±‌

π

2

or ‌‌y=sin(±‌

π

2

) or y=±1
Therefore, ordered pair is (cos‌

π2

4

,±1).