as we know that -
[tex]\sf{Cos(A + B) \: = \: CosA CosB \: - \: SinA SinB}[/tex]
[tex]\: \: \: \: \: \: \: \: \sf{(A + B) \: = \: Cos^{-1}\bigg(CosA CosB - SinA SinB\bigg) \: -(1)}[/tex]
let -
[tex]\: \: \: \: \: \: \sf{CosA \: = \: x}[/tex]
[tex]\: \: \: \: \: \: \sf{CosB \: = \: y}[/tex]
So -
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \sf{A \: = \: Cos^{-1}x}[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \sf{B \: = \: Cos^{-1}y}[/tex]
and
[tex]\: \: \: \: \: \: \sf{SinA \: = \: \sqrt{(1 - Cos^{2}A)} = \sqrt{1 - x^{2}}}[/tex]
[tex]\: \: \: \: \: \: \sf{SinB \: = \: \sqrt{(1 - Cos^{2}B)} = \sqrt{1 - y^{2}}}[/tex]
Now -
On putting above values in eqⁿ(1) -
[tex]\sf{Cos^{-1}x + Cos^{-1}y \: = \: Cos^{-1}\bigg(xy - \sqrt{1 - x^{2}} \sqrt{1 - y^{2}}}[/tex]
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Answers & Comments
as we know that -
[tex]\sf{Cos(A + B) \: = \: CosA CosB \: - \: SinA SinB}[/tex]
[tex]\: \: \: \: \: \: \: \: \sf{(A + B) \: = \: Cos^{-1}\bigg(CosA CosB - SinA SinB\bigg) \: -(1)}[/tex]
let -
[tex]\: \: \: \: \: \: \sf{CosA \: = \: x}[/tex]
[tex]\: \: \: \: \: \: \sf{CosB \: = \: y}[/tex]
So -
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \sf{A \: = \: Cos^{-1}x}[/tex]
[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \sf{B \: = \: Cos^{-1}y}[/tex]
and
[tex]\: \: \: \: \: \: \sf{SinA \: = \: \sqrt{(1 - Cos^{2}A)} = \sqrt{1 - x^{2}}}[/tex]
[tex]\: \: \: \: \: \: \sf{SinB \: = \: \sqrt{(1 - Cos^{2}B)} = \sqrt{1 - y^{2}}}[/tex]
Now -
On putting above values in eqⁿ(1) -
[tex]\sf{Cos^{-1}x + Cos^{-1}y \: = \: Cos^{-1}\bigg(xy - \sqrt{1 - x^{2}} \sqrt{1 - y^{2}}}[/tex]