Question

cos (x+y).cos (x-y) = cos²y - sin²x​

Answer 1

${\boxed{\boxed{\red{\mathcal{ANSWER:}}}}}$

Given:

• Cos(x + y)Cos(x - y) = Cos²y - Sin²x

Formula used:

• Cos(a + b) = Cos a.Cos b + Sin a.Sin b
• Cos(a - b) = Cos a.Cos b - Sin a.Sin b

${\boxed{\boxed{\red{\mathcal{SOLUTION:}}}}}$

We will choose LHS part,

$\implies$ Cos(x + y) Cos(x-y)

$\implies$ (Cos x Cos y + Sin x Sin y)(Cos x Cos y - Sin x Sin y)

Now, by using (a + b) (a - b) = a² - b²

$\implies$ (Cos x Cos y)² - (Sin x Sin y)²

$\implies$ Cos² x Cos² y - Sin² x Sin² y

$\implies$ Cos² y(1 - Sin² x) - (1 - Cos² y)Sin² x

$\implies$ Cos² y - Cos² y Sin² x - Sin² yx + Cos² y Sin² x

$\implies$ Cos²y - Sin²x

$\implies$ RHS

Hence Proved!!!

Answer 2

Given:

Cos(x + y)Cos(x - y) = Cos²y - Sin²x

Formula used:

Cos(a + b) = Cos a.Cos b + Sin a.Sin b

Cos(a - b) = Cos a.Cos b - Sin a.Sin b

Solution:

We will choose LHS part ,

⟹ Cos(x + y) Cos(x-y)

⟹ (Cos x Cos y + Sin x Sin y)(Cos x Cos y - Sin x Sin y)

Now, by using (a + b) (a - b) = a² - b²

⟹ (Cos x Cos y)² - (Sin x Sin y)²

⟹ Cos² x Cos² y - Sin² x Sin² y

⟹ Cos² y(1 - Sin² x) - (1 - Cos² y)Sin² x

⟹ Cos² y - Cos² y Sin² x - Sin² yx + Cos² y Sin² x

⟹ Cos²y - Sin²x

⟹ RHS

Hence proved!!!