Question

(sina+cosa)^2-(sina-cosa)^2​

Answer 1

★ To Find :

We have to find the value of (Sin a+ Cos a)² - (Sin a - Cos a)²

$\rule{200}{1}$

★ Solution :

We know that,

$\Large{\star{\boxed{\sf{(a + b)^2 = a^2 + b^2 + 2ab}}}} \\ \Large{\star{\boxed{\sf{(a - b)^2 = a^2 + b^2 - 2ab}}}}$

Where,

• a is Sin a
• b ia Cos a

___________[Put Values]

$\tt{\implies Sin^2a + cos^2a + 2(Sin a)(Cos a) - \bigg( Sin^2a + cos^2a - 2(Sin a)(Cos a) \bigg)} \\ \\ \tt{\implies \cancel{Sin^2a} + \cancel{cos^2a} + 2Sin \: a.Cos \: a \cancel{- Sin^2a} \cancel{ - cos^2a} + 2Sin \ :a .Cos \: a} \\ \\ \tt{\implies 4Sin \: a . Cos \:a}$