Question

Answer me back....………

Answer 1

⭐Solutions⭐

✏Prove that's

(sin⁴A + Cos⁴ A ) = 1-2 sin²A Cos²A

Now,

Take LHS,

= Sin⁴A + Cos ⁴A

= (Sin ²A)² + (Cos²A)²

= (Sin² A + Cos ²A)² - 2Sin ²A Cos²A

We know

[ ♠ (a²+b²)= (a+b)² - 2ab]

[ ♠ Sin ²A + Cos²A = 1]

= (1)² - 2Sin²A Cos²A

= 1- 2 sin² A Cos² A

= RHS

✏Thats Proved .

✒hopes its help's u.

Answer 2

Answer:

$\boxed{\text{sin}^{4}(A)+\text{cos}^{4}(A)=1-\left(2\cdot\text{sin}^{2}A\cdot\text{cos}^{2}A\right)}}$

Step-by-step explanation:

Solve the left side of the given equation as shown below:

$\begin{aligned}\text{sin}^{4}(A)+\text{cos}^{4}(A)&=\left(\text{sin}^{2}(A)\right)^{2}+\left(\text{cos}^{2}(A)\right)^{2}\\&=\left(\text{sin}^{2}(A)+\text{cos}^{2}(A)\right)^{2}-\left(2\cdot \text{sin}^{2}(A)\cdot \text{cos}^{2}(A)\right)\\&=1-\left(2\cdot \text{sin}^{2}(A)\cdot \text{cos}^{2}(A)\right)\end{aligned}$

From the above calculation it is proved that $\boxed{\text{sin}^{4}(A)+\text{cos}^{4}(A)=1-\left(2\cdot\text{sin}^{2}A\cdot\text{cos}^{2}A\right)}}$