Question

give me the answer of the 2nd question​

Answer 1

Here is your answer

sin+cos=1

sq both the sides

sin^2+cos^2+2sincos=1

1+2sincos=1

sincos=0

Thanks.....

Answer 2

Answer: 0

Step-by-step explanation:

We are given sinθ+cosθ=1 ----------equation 1

We need to find sinθ X cosθ

Now, squaring on both sides of equation 1, we get

(sinθ + cosθ)Λ2 = (1)Λ2 ----------------equation 2

Using the property, (a+b)Λ2 = (a)Λ2 + (b)Λ2 + 2 (a X b)

Here a = sinθ and b = cosθ, now equation 2 becomes,

(sinθ)Λ2 + (cosθ)Λ2 + 2 (sinθ X cosθ) = (1)Λ2

Using property, (sinθ)Λ2 +(cosθ)Λ2 = 1 , in the above equation, we get

1 + 2 (sinθ X cosθ) = 1

2 (sinθ X cosθ) = 1-1

2 (sinθ X cosθ) = 0

(sinθ X cosθ) = 0 ------------------------------------ Answer