Question

sin (180° + x). cos (270° - x) + cos (180° - x). sin (270° + x) = 1
​

Answer 1

To prove:

$sin (180^\circ + x). cos (270^\circ - x) + cos (180^\circ - x). sin (270^\circ + x) = 1$

Solution:

Let us consider the Left Hand Side first.

$sin(180^\circ+\theta)$ will be the angle in 3rd quadrant, sine is negative in 3rd quadrant so result will be negative.

$\therefore sin(180^\circ+\theta) = -sin\theta$

$cos(270^\circ-\theta)$ will be the angle in 3rd quadrant, cosine is negative in 3rd quadrant so result will be negative.

$\therefore cos(270^\circ-\theta) = -sin\theta$

$cos(180^\circ-\theta)$ will be the angle in 2nd quadrant, cosine is negative in 2nd quadrant so result will be negative.

$\therefore cos(180^\circ-\theta) = -cos\theta$

$sin(270^\circ+\theta)$ will be the angle in 4th quadrant, sine is negative in 4th quadrant so result will be negative.

$\therefore sin(270^\circ+\theta) = -cos\theta$

Putting the values in LHS:

$sin (180^\circ + x). cos (270^\circ - x) + cos (180^\circ - x). sin (270^\circ + x)\\\Rightarrow (-sinx)(-sinx)+(-cosx)(-cosx)\\\Rightarrow sin^2x+cos^2x = 1\ \ \ \ \ \ (\because sin^2\theta+cos^2\theta = 1)$

Hence Proved that LHS = RHS

OR

$sin (180^\circ + x). cos (270^\circ - x) + cos (180^\circ - x). sin (270^\circ + x) = 1$

Answer 2

Given:

sin (180° + x). cos (270° - x) + cos (180° - x). sin (270° + x) = 1

To find:

​To prove: sin (180° + x). cos (270° - x) + cos (180° - x). sin (270° + x) = 1

Solution:

From given, we have,

sin (180° + x). cos (270° - x) + cos (180° - x). sin (270° + x) = 1

Now consider L.H.S:

sin (180° + x). cos (270° - x) + cos (180° - x). sin (270° + x)

we use the properties:

sin (180° + x) = -sin x

cos (270° - x)  = -sin x

cos (180° - x) = -cos x

sin (270° + x) =  -cos x

so, we get,

= (-sin x) × (-sin x) + (-cos x) (-cos x)

= sin² x + cos² x

= 1

= R.H.S

Hence proved.