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Question
In ∆ABC, prove that `("b" - "c")^2 cos^2 ("A"/2) + ("b" + "c")^2 sin^2 ("A"/2)` = a2
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Solution
L.H.S. = `("b" - "c")^2 cos^2 "A"/2 + ("b" + "c")^2 sin^2 "A"/2`
= `("b"^2 + "c"^2 - 2"bc") cos^2 "a"/2 + ("b"^2 + "c"^2 + 2"bc") sin^2 "A"/2`
= `("b"^2 + "c"^2) cos^2 "A"/2 - 2"bc" cos^2 "A"/2 + ("b"^2 + "c"^2) sin^2 "A"/2 + 2"bc" sin^2 "A"/2`
=`("b"^2 + "c"^2) (cos^2 "A"/2 + sin^2 "A"/2) - 2"bc" (cos^2 "A"/2 - sin^2 "A"/2)`
= (b2 + c2)(1) − 2bc cos A .......[∵ cos2θ − sin2θ = cos 2θ]
= b2 + c2 − 2bc cos A
= a2 .......[By cosine rule]
= R.H.S.
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