1.

(sin ø- cos ø)²+(sin ø- cos ø)²=​

Answer» ONG>Answer:

\displaystyle \sf{ \red{ \bold{ Find}}}

{ \sf{➣( {sinθ - cosθ)}^{2} + {(sin θ- cosθ)}^{2} }}

{ \bold{ \blue{Solution}}}

{ \sf{⟶ {(sinθ - cosθ)}^{2} + {(sinθ - cosθ)}^{2} }}

From, ALGEBRAIC formula.....

{ \boxed{ \bold{ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab}}}

{ \sf{⟶ {sin}^{2}θ + {cos}^{2} θ - 2sinθcosθ + {sin}^{2} θ + {cos}^{2} θ - 2sinθcosθ}}

From trigonometry identity:-

{ \boxed{ {sin}^{2} + {cos}^{2} = 1 }}

{ \sf{⟶1 - 2sinθcosθ + 1 - 2sinθcosθ}}

{ \sf{⟶2 - 4sinθcosθ}}

Therefore, (sin θ- cos θ)²+(sin θ- cos θ)²=2-4sinθcosθ


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