Question

$\red {10}$
Find the distance between origin and
(sinA+cosA , sinA-cosA ).​

Answer 1

√2 is The distance between the points ( cosA, sinA ) and ( -sinA, cosA )

Step-by-step explanation:

we need to find The distance between the points ( cosA, sinA ) and ( -sinA, cosA )

Distance between Two Points A (x₁ , y₁) & B (x₂ , y₂)

is given by √(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = CosA y₁ = SinA

x₂ = -SinA y₂ = CosA

Distance = √(-SinA - CosA)² + (CosA - SinA)²

=√Sin²A + Cos²A + 2SinACosA + Cos²A + Sin²A - 2SinACosA

as we know that Sin²A + Cos²A = 1

= √1 + 2SinACosA +1 - 2SinACosA

= √2

The distance between the points ( cosA, sinA ) and ( -sinA, cosA ) is √2

Learn more:

23. Find the distance between the points:(3,-2) and (15,3) - Brainly.in

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

To Find :

Distance between origin and (SinA + CosA , SinA - CosA)

______________________________

Solution :

We know that coordinates of the origin are (0,0). Take A(0,0)

And the coordinates of other points are (SinA + CosA , SinA - CosA),Take B(SinA + CosA, SinA - CosA)

Use distance formula :

$\Large{\underline{\boxed{\sf{|AB| \: = \: \sqrt{(x_{2} \: - \: x_1)^2 \: + \: (y_2 \: - \: y_1)^2}}}}}$

Where,

x1 = 0

y1 = 0

x2 = SinA + CosA

y2 = SinA - CosA

____________[Put Values]

⇒|AB| = √(SinA + CosA- 0)² + (SinA - CosA - 0)²

⇒|AB| = √(SinA + CosA)² + (SinA - CosA)²

⇒|AB| = √(Sin²A + Cos²A + 2SinACosA + Sin²A + Cos²A - 2SinACosA)

⇒|AB| = √(1 + 2SinACosA) + (1 - 2SinACosA)

⇒|AB| = √2 + 2SinACosA - 2SinACosA

⇒|AB| = √2

$\Large{\boxed{\boxed{\sf{|AB| \: = \: \sqrt{2} \: units }}}}$

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Important Formulas :

• Sin²A + Cos²A = 1

• Cosec²A - Cot²A = 1

• Sec²A - Tan²A = 1

• SinA/CosA = TanA

• CosA/SinA = CotA

#answerwithquality

#BAL