Sin-1x+Sin-1y+Sin-1z=3(pi)/2 then find x+y+z=?
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Since maximum value of sin angle is 1 i.e., thus if we choose
sin⁻¹x=sin⁻¹y=sin⁻¹z=π/2 [∵, sin(π/2)=1] that will satisfy the given equation as follows:
sin⁻¹x+sin⁻¹y+sin⁻¹z=π/2+π/2+π/2=3π/2
∴, x=y=z=sinπ/2=1
∴, x+y+z=1+1+1=3 Ans.
sin⁻¹x=sin⁻¹y=sin⁻¹z=π/2 [∵, sin(π/2)=1] that will satisfy the given equation as follows:
sin⁻¹x+sin⁻¹y+sin⁻¹z=π/2+π/2+π/2=3π/2
∴, x=y=z=sinπ/2=1
∴, x+y+z=1+1+1=3 Ans.
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