Answer:

[tex]option \: - \: ( \: 3 \: ) \frac{1}{2} ( {a}^{2} + {b}^{2} - 2)[/tex]

Step-by-step explanation:

[tex]hello \: samira[/tex]

[tex]hope \: this \: helps \: you.[/tex]

[tex](⁠ʘ⁠ᴗ⁠ʘ⁠✿⁠)[/tex]

Verified answer

Answer:

Let

sin x + sin y = a ------------(i)

And

cos x + cos y = b ---------(ii)

Squaring equation (i) on both side we get

=> (sinx + siny)² = a²

=> sin²x + sin²y + 2sinxsiny = a² ----(iii)

Now squaring equation (ii) we get

=>( cosx + cosy )² = b²

=> cos²x + cos²y + 2cosxcosy = b² ---(iv)

Now adding equation (iii) & (iv ) we get

[tex] \sf\dashrightarrow {sin}^{2}x + {sin}^{2}y + 2sinxsiny + {cos}^{2}x + {cos}^{2} y + 2cosxcosy = {a}^{2} + {b}^{2} \\ \\ \sf\dashrightarrow \: {sin}^{2}x + {cos}^{2}x + {sin}^{2} y + {cos}^{2}y + 2sinxsiny + 2cosxcosy = {a}^{2} + {b}^{2} \\ \\ \sf\dashrightarrow \: 1 + 1 + 2(sinxsiny + cosxcosy) = {a}^{2} + {b}^{2} \\ \\ \sf\dashrightarrow \: 2 + 2(sinxsiny + cosxcosy) = {a}^{2} + {b}^{2} \\ \\ \sf\dashrightarrow \: 2 + 2 \times cos(x - y) = {a}^{2} + {b}^{2} \\ \\ \sf\dashrightarrow \: 2 \times cos(x - y) = {a}^{2} + {b}^{2} - 2 \\ \\ \sf\dashrightarrow \: cos(x - y) = \dfrac{ {a}^{2} + {b}^{2} - 2 }{2} [/tex]

• Hence , the required option is (c) ½ ( a² + b² - 2 )

Formula Used here

[tex]\bullet\sf\: cos(x - y) = sinxsiny + cosxcosy \\ \\ \bullet \: \sf \: {a}^{2} + {b}^{2} + 2ab = (a + b)^{2}[/tex]