[tex]\large\underline{\sf{Solution-}}[/tex]

Consider,

[tex]\sf \: cos\left(\dfrac{A + B}{2} \right)cos\left(\dfrac{A - B}{2} \right) \\ [/tex]

can be rewritten as

[tex]\sf \: = \: cos\left(\dfrac{A}{2} + \dfrac{B}{2} \right)cos\left(\dfrac{A}{2} - \dfrac{B}{2} \right) \\ [/tex]

We know,

[tex]\boxed{\begin{aligned}& \qquad \:\sf \: cos(x + y) = cosxcosy - sinxsiny \qquad \: \\ \\& \qquad \:\sf \:cos(x - y) = cosxcosy + sinxsiny \end{aligned}} \qquad \: \\ [/tex]

So, using these Identities, the above expression can be rewritten as

[tex]\sf \: = \bigg(cos\dfrac{A}{2} cos\dfrac{B}{2} - sin\dfrac{A}{2}sin\dfrac{B}{2} \bigg) \bigg( cos\dfrac{A}{2}cos\dfrac{B}{2} + sin\dfrac{A}{2}sin\dfrac{B}{2}\bigg) \\ [/tex]

[tex]\sf \: = \: {\bigg( cos\dfrac{A}{2}cos\dfrac{B}{2}\bigg) }^{2} - {\bigg( sin\dfrac{A}{2}sin\dfrac{B}{2}\bigg) }^{2} \\ [/tex]

[tex]\sf \: = \: {cos}^{2}\dfrac{A}{2} {cos}^{2}\dfrac{B}{2} - {sin}^{2}\dfrac{A}{2} {sin}^{2}\dfrac{B}{2} \\ [/tex]

[tex]\sf \: = \: {cos}^{2}\dfrac{A}{2} \left(1 - {sin}^{2}\dfrac{B}{2}\right) - \left(1 - {cos}^{2}\dfrac{A}{2}\right) {sin}^{2}\dfrac{B}{2} \\ [/tex]

[tex]\sf \: = \: {cos}^{2}\dfrac{A}{2} - {cos}^{2}\dfrac{A}{2} {sin}^{2}\dfrac{B}{2} - {sin}^{2}\dfrac{B}{2} + {sin}^{2}\dfrac{B}{2} {cos}^{2}\dfrac{A}{2} \\ [/tex]

[tex]\sf \: = \: {cos}^{2}\dfrac{A}{2} - {sin}^{2}\dfrac{B}{2} \\ [/tex]

Hence,

[tex]\boxed{\sf \: \sf \: cos\left(\dfrac{A + B}{2} \right)cos\left(\dfrac{A - B}{2} \right) = {cos}^{2}\dfrac{A}{2} - {sin}^{2}\dfrac{B}{2} \: } \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(x - y) = sinx \: cosy \: - \: siny \: cosx}\\ \\ \bigstar \: \bf{sin(x + y) = sinx \: cosy \: + \: siny \: cosx}\\ \\ \bigstar \: \bf{cos(x + y) = cosx \: cosy \: - \: sinx \: siny}\\ \\ \bigstar \: \bf{cos(x - y) = cosx \: cosy \:+\: siny \: sinx}\\ \\ \bigstar \: \bf{tan(x + y) = \dfrac{tanx + tany}{1 - tanx \: tany} }\\ \\ \bigstar \: \bf{tan(x - y) = \dfrac{tanx - tany}{1 + tanx \: tany} }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]

Answer:

To prove the identity:

cos(A+B)/2 * cos(A-B)/2 = cos^2A/2 - sin^2B/2

We start by using the identity:

cos 2θ = cos^2θ - sin^2θ

Let's replace θ with (A+B)/2:

cos (A+B) = cos^2(A+B)/2 - sin^2(A+B)/2

Now, let's replace θ with (A-B)/2:

cos (A-B) = cos^2(A-B)/2 - sin^2(A-B)/2

We can rearrange these equations to solve for cos^2(A+B)/2 and cos^2(A-B)/2:

cos^2(A+B)/2 = cos(A+B) + sin^2(A+B)/2

cos^2(A-B)/2 = cos(A-B) + sin^2(A-B)/2

Substituting these expressions into the left-hand side of the original identity, we get:

cos(A+B)/2 * cos(A-B)/2

= (cos(A+B) + sin^2(A+B)/2) * (cos(A-B) + sin^2(A-B)/2) (Substituting the expressions for cos^2(A+B)/2 and cos^2(A-B)/2)

= cos(A+B)cos(A-B) + (cos(A+B)sin^2(A-B)/2 + cos(A-B)sin^2(A+B)/2 + sin^2(A+B)/2sin^2(A-B)/2)

= cos^2A/2 - sin^2B/2 (Substituting the trigonometric identities for cos(A+B) and cos(A-B), and using the identity cos^2θ + sin^2θ = 1)

Therefore, we have proven the identity:

cos(A+B)/2 * cos(A-B)/2 = cos^2A/2 - sin^2B/2