Answer

Verified

483.9k+ views

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.

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.

Recently Updated Pages

what is the correct chronological order of the following class 10 social science CBSE

Which of the following was not the actual cause for class 10 social science CBSE

Which of the following statements is not correct A class 10 social science CBSE

Which of the following leaders was not present in the class 10 social science CBSE

Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE

Which one of the following places is not covered by class 10 social science CBSE

Trending doubts

Which places in India experience sunrise first and class 9 social science CBSE

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How do you graph the function fx 4x class 9 maths CBSE

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE