Question

Evaluate (sinx - cosx)^2 + (sinx + cosx)^2​

Answer 1

(sinx - cosx)² + (sinx + cosx)²

=> (sin²x - 2sinxcosx + cos²x) + (sin²x + 2sinxcosx + cos²x)

=> 1 - 2sinxcosx + 1 + 2sinxcosx

=> 2

{ As we know that : sin²A + cos²A = 1 }

I hope it will be helpful for you ✌️✌️

Mark it as brainliest and follow me ☺️☺️

Answer 2

$\huge\boxed{\fcolorbox{cyan}{grey}{Solution:-}}$.

➡Given here,

↪(Sinx-Cosx)²+(Sinx+Cosx)².

(Sinx-Cosx)²+(Sinx+Cosx)².= (Sin²x+cos²x-2sinx cosx)+(sin²x+cos²x+2sinx cosx)

(Sinx-Cosx)²+(Sinx+Cosx)².= (Sin²x+cos²x-2sinx cosx)+(sin²x+cos²x+2sinx cosx)= 2(sin²x+cos²x)

(Sinx-Cosx)²+(Sinx+Cosx)².= (Sin²x+cos²x-2sinx cosx)+(sin²x+cos²x+2sinx cosx)= 2(sin²x+cos²x)= 2(1)

(Sinx-Cosx)²+(Sinx+Cosx)².= (Sin²x+cos²x-2sinx cosx)+(sin²x+cos²x+2sinx cosx)= 2(sin²x+cos²x)= 2(1)=1.

➡We know .

↪Sin²x+Cos²x=1.

$\huge\boxed{\fcolorbox{cyan}{grey}{</u></em></strong><strong><em><u>H</u></em></strong><strong><em><u>o</u></em></strong><strong><em><u>p</u></em></strong><strong><em><u>e</u></em></strong><strong><em><u>s</u></em></strong><strong><em><u> </u></em></strong><strong><em><u>i</u></em></strong><strong><em><u>t</u></em></strong><strong><em><u>s</u></em></strong><strong><em><u> </u></em></strong><strong><em><u>h</u></em></strong><strong><em><u>e</u></em></strong><strong><em><u>l</u></em></strong><strong><em><u>p</u></em></strong><strong><em><u>s</u></em></strong><strong><em><u> </u></em></strong><strong><em><u>u</u></em></strong><strong><em><u>:-}}$.