Question

cos(π/2-x).cos(π/2-y)-sin(π/2-x).(sinπ/2-y) = cos(x+y)​

Answer 1

EXPLANATION.

⇒ cos(π/2 - x). cos(π/2 - y) - sin(π/2 - x). sin(π/2 - y).

As we know that,

Formula of :

⇒ cos(π/2 - θ) = sinθ.

⇒ sin(π/2 - θ) = cosθ.

Using this formula in the equation, we get.

⇒ sin(x). sin(y) - cos(x). cos(y).

As we know that,

Formula of :

⇒ cos(x + y) = cos(x). cos(y) - sin(x). sin(y).

We can write equation as,

⇒ - [cos(x).cos(y) - sin(x).sin(y)].

⇒ - cos(x + y).

                                                                                                                 

MORE INFORMATION.

(1) sin2θ = 2sinθcosθ = 2tanθ/1 + tan²θ.

(2) cos2θ = cos²θ - sin²θ = 2cos²θ - 1 = 1 - 2sin²θ = 1 - tan²θ/1 + tan²θ.

(3) tan2θ = 2tanθ/1 - tan²θ.

(4) sin3θ = 3sinθ - 4sin³θ.

(5) cos3θ = 4cos³θ - 3cosθ.

(6) tan3θ = 3tanθ - tan³θ/1 - 3tan²θ.

Answer 2

$\sf\green {\tt{Answer}}$

$\sf\purple{ \tt{ \cos( \frac{\pi}{2} - x) }}$

$\sf\pink{ \cos( \frac{\pi}{2} - y) - \sin( \frac{\pi}{2} - y) = \cos(x + y) }$

{\implies -cos(x+y)}