Given, y=[∣sinx∣+∣cosx∣] and x2+y2=10
We know that (∣sinx∣+∣cosx∣)∈[1,2] ∴y=1
The point of intersection of given curve is x2+12=10 ⇒x2=9 ⇒x=±3 ∴ Point of intersection is (±3,1)
Now, x2+y2=10 ⇒2x+2ydxdy=0 ⇒dxdy=−yx
At point (−3,1) dxdy=13=3 ⇒m1=3
Slope of line y=1 is m2=0 ∴ Angle between two curves is tanθ=1+m1m2m1−m2=3 ⇒θ=tan−1(3)