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# Prove the following trigonometric equation :${{\text{sin}}^2}{\text{A = co}}{{\text{s}}^2}({\text{A - B}}) + {\cos ^2}{\text{B}} - 2\cos ({\text{A - B}})\cos {\text{AcosB}}$  Answer Verified
Hint: In order to solve such types of problems, we must keep one thing in our mind that how can we arrange the terms so that we can apply available trigonometric formulas then expression will automatically start to get reduced.

Complete step-by-step answer:
$\Rightarrow {\text{si}}{{\text{n}}^2}{\text{A = co}}{{\text{s}}^2}({\text{A - B}}) + {\cos ^2}{\text{B}} - 2\cos ({\text{A - B}})\cos {\text{AcosB}}$
On writing ${\text{co}}{{\text{s}}^2}{\text{B}}$ term first in RHS as shown below, because through such type rearrangement we can proceed to solve by taking ${\text{cos}}({\text{A - B}})$ term common
$\Rightarrow {\text{RHS = co}}{{\text{s}}^2}({\text{A - B}}) + {\cos ^2}{\text{B}} - 2\cos ({\text{A - B}})\cos {\text{AcosB}}$
On taking ${\text{cos}}({\text{A - B}})$ term common,
$\Rightarrow {\text{ }}{\cos ^2}{\text{B + cos}}({\text{A - B}})\left( {(\cos ({\text{A - B}}) - 2\cos {\text{AcosB)}}} \right)$ --- (1)
We know that.
$\Rightarrow {\text{ cos}}({\text{A - B}}) = {\text{ cosAcosB + sinAsinB}}$
So on putting the value of ${\text{cos}}({\text{A - B}})$ in expression (1)
$\Rightarrow {\cos ^2}B{\text{ + cos}}({\text{A - B}})\left( {(\cos {\text{AcosB + sinAsinB}} - 2\cos {\text{AcosB)}}} \right)$
On subtracting ${\text{cos}}({\text{A)cos (B)}}$ term we get
$\Rightarrow {\cos ^2}B{\text{ + cos}}({\text{A - B}})\left( {({\text{sinAsinB}} - \cos {\text{AcosB)}}} \right)$
On rearranging minus sign, try to make any formula of cos
$\Rightarrow {\cos ^2}B{\text{ + cos}}({\text{A - B}})\left( {( - \cos {\text{AcosB + sinAsinB)}}} \right)$
On taking minus sign common from 2nd bracket
$\Rightarrow {\cos ^2}B{\text{ + cos}}({\text{A - B}})\left( { - (\cos {\text{AcosB - sinAsinB)}}} \right)$
We can take this minus sign out from 2nd bracket
$\Rightarrow {\cos ^2}B{\text{ - cos}}({\text{A - B}})\left( {\cos {\text{AcosB - sinAsinB}}} \right)$ --- (2)
We know the formula of ${\text{cos}}({\text{A + B}})$ so here we can use that
${\text{ cos}}({\text{A + B}}) = {\text{ cosAcosB - sinAsinB}}$
So using the formula of ${\text{cos}}({\text{A + B}})$ in expression (2)
$\Rightarrow {\cos ^2}B{\text{ - cos}}(A - B)(\cos ({\text{A + B}}){\text{)}}$ --- (3)
We know that.
${\text{ cos}}({\text{A - B}}) = {\text{ cosAcosB + sinAsinB}}$
${\text{ cos}}({\text{A + B}}) = {\text{ cosAcosB - sinAsinB}}$
On using above results of ${\text{cos}}({\text{A - B}}){\text{ & cos}}({\text{A - B}})$ in expression (3)
$\Rightarrow {\cos ^2}B{\text{ - }}\left( {{\text{(cosAcosB + sinAsinB)}}(\cos {\text{AcosB - sinAsinB)}}} \right)$
In algebra, there is a formula known as the Difference of two squares:$({{\text{m}}^2}{\text{ - }}{{\text{n}}^2}) = ({\text{m + n}})({\text{m - n}})$
Here, ${\text{m = cosAcosB}}$ ${\text{ & n = sinAsinB}}$
So on using Difference of two squares formula
$\Rightarrow {\cos ^2}B{\text{ - }}\left( {{\text{(co}}{{\text{s}}^{^2}}{\text{Aco}}{{\text{s}}^2}{\text{B - si}}{{\text{n}}^2}{\text{Asi}}{{\text{n}}^2}{\text{B}}} \right)$
On further solving
$\Rightarrow {\cos ^2}B{\text{ - co}}{{\text{s}}^{^2}}{\text{Aco}}{{\text{s}}^2}{\text{B + si}}{{\text{n}}^2}{\text{Asi}}{{\text{n}}^2}{\text{B}}$
On taking the ${\text{co}}{{\text{s}}^2}({\text{B}})$ term common
$\Rightarrow {\cos ^2}B{\text{ (1 - co}}{{\text{s}}^{^2}}{\text{A) + si}}{{\text{n}}^2}{\text{Asi}}{{\text{n}}^2}{\text{B}}$
We know the formula of ${\text{si}}{{\text{n}}^2}{\text{A + co}}{{\text{s}}^2}{\text{A = 1 }}$ so here we can use
${\text{ (1 - co}}{{\text{s}}^{^2}}{\text{A) = si}}{{\text{n}}^2}{\text{A}}$ in above expression
$\Rightarrow {\cos ^2}B{\text{ (si}}{{\text{n}}^{^2}}{\text{A) + si}}{{\text{n}}^2}{\text{Asi}}{{\text{n}}^2}{\text{B}}$
On taking the ${\sin ^2}({\text{A}})$ term common
$\Rightarrow {\text{(si}}{{\text{n}}^{^2}}{\text{A) }}\left( {{{\cos }^2}B{\text{ + si}}{{\text{n}}^2}{\text{B}}} \right)$
Again using ${\text{si}}{{\text{n}}^2}{\text{B + co}}{{\text{s}}^2}{\text{B = 1 }}$
$\Rightarrow {\text{(si}}{{\text{n}}^{^2}}{\text{A) }} \times \left( 1 \right)$
$\Rightarrow {\text{si}}{{\text{n}}^{^2}}{\text{A = LHS}}$

Note: Whenever we face such a type of problem always remember the trigonometry identities which are written above then simplify the given statements using these identities. we will get the required answer.
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Trigonometric Equations  Trigonometric Functions  Trigonometric Ratios  Trigonometric Identities  Trigonometry Values  CBSE Class 11 Maths Chapter 3 - Trigonometric Functions Formulas  Inverse Trigonometric Functions  Trigonometric Identities - Class 10  Integration of Trigonometric Functions  Trigonometric Ratios of Complementary Angles  