Answer: Verified.
Step-by-step explanation: The given identity involving trigonometric ratios is
[tex]\cos(x+y)\cos(x-y)=\cos^2x-\sin^2y.[/tex]
We have
[tex]L.H.S.\\\\=\cos(x+y)\cos(x-y)\\\\=\left(\cos x\cos y-\sin x\sin y\right)\left(\cos x\cos y+\sin x\sin y\right)\\\\=\left(\cos x\cos y\right)^2-\left(\sin x\sin y\right)^2\\\\=\cos^2x\cos^2y-\sin^2x\sin^2y\\\\=\cos^2x(1-\sin^2y)-(1-\cos^2x)\sin^2y\\\\=\cos^2x-\cos^2x\sin^2y-\sin^2y+\cos^2x\sin^2y\\\\=\cos^2x-\sin^2y\\\\=R.H.S.[/tex]
Therefore, L.H.S. = R.H.S.
Thus, the given identity is verified.
Answer:
cos(x+y)sin(x-y) = cos²x-sin²y ⇒ L.H.S = R.H.S
Step-by-step explanation:
L.H.S:
∵ cos(x+y)cos(x-y) = (cosxcosy - sinxsiny)(cosxcosy + sinxsiny)
Multiply the brackets ⇒ the answer is difference of two squares
∴ = cos²xcos²y - sin²xsin²y
∵ cos²y = 1 - sin²y and sin²x = 1 - cos²x
∴ cos²xcos²y - sin²xsin²y = cos²x (1 - sin²y) - (1 -cos²x) sin²y
∴ = cos²x - cos²xsin²y - sin²y + cos²xsin²y
∵ -cos²x sin²y will cancel cos²xsin²y ⇒ same terms with different sign
∴ = cos²x - sin²y = R.H.S
