Question

Prove that,

(i) sin(A+B)sin(A-B)=cos²B- cos²A

(ii)cos(A+B)cos(A-B)=cos²B - sin²A​

Answer 1

★ Solution No.(i) ★

Taking L.H.S ,

sin(A+B)sin(A-B)

=(sinA cosB+cosA sinB).(sinA cosB - cosA sinB)

= sin²A cos²B - cos²A sin²B

= sin²A(1-sin²B)-(1-sin²A)sin²B

= sin²A-sin²A sin²B - sin²B + sin²A sin²B

= sin²A - sin²B

= (1-cos²A) - (1-cos²B)

=1-cos²A-1 +cos²B

= cos²B - cos²A [ Proved ]

★ Solution No.(ii) ★

Taking L.H.S ,

cos(A+B)cos(A-B)

=(cosA cosB + sinA sinB).(cosA cosB - sinA sinB)

= cos²A cos²B - sin²A sin²B

= cos²A(1-sin²B)-(1-cos²A)sin²B

= cos²A-cos²A sin²B - sin²B + cos²A sin²B

= cos²A - sin²B

= (1-sin²A) - (1-cos²B)

=1-sin²A-1 +cos²B

= cos²B - sin²A [ Proved ]

______________________

Some formulas :-

★sin(A+B)=sinA cosB + cosA sinB

★sin(A-B) = sinA cosB - cosA sinB

★cos(A+B)=cosA cosB - sinA sinB

★cos(A-B) =cosA cosB + sinA sinB

____________________

Answer 2

$\large\underline\mathrm{Solution}$

$\huge \mathfrak{LHS:-}$

$\implies$ sin(A+B)sin(A-B) = (sinA cosB - cos A sin B)(sinA cosB - cosA sinB)

$\implies$ sin²A cos²B - cos²A sin²B

$\implies$ sin²A (1 - sin²B) - (1 - sin²A) sin²B

$\implies$ sin²A - sin²B sin²B - sin²B + sin²A sin²B

$\implies$ sin²A - sin²B

$\implies$ (1 - cos²A)(1 - cos²B)

$\implies$ 1 - cos²A - 1 + cos²B

$\implies$ cos²B - cos²A

$\huge \mathfrak{proved:-}$

$\huge \mathfrak{LHS:-}$

$\implies$ cos(A + B)cos(A - B) = (cosA cosB + sinA sinB)(cosA cosB - sinA sinB)

$\implies$ cos²A cos²B - sin²A sin²B

$\implies$ cos²A(1 + sin²B) - (1 - cos ²A)sin²B

$\implies$ cos²A - cos²A sin²B - sin²B + cos²A sin²B

$\implies$ cos²A - sin²B

$\implies$ (1 - sin²A) - (1 - cos²B)

$\implies$ 1 - sin²A - 1 + cos²B

$\implies$ cos²B - sin²A

$\huge \mathfrak{proved:-}$

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