The sides a,b,c of △ABC, are in A.P.If cosα=ab+c,cosβ=bc+a,cosγ=ca+b then tan2α2+tan2γ2=
A
1
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B
12
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C
13
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D
23
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Solution
The correct option is D23 Given:a,b,c are in A.P ⇒2b=a+c cosα1=ab+c By componendo-dividendo rule, we have 1−cosα1+cosα=b+c−ab+c+a ⇒tan2α2=b+c−a3b where a+c=2b cosγ1=ca+b By componendo-dividendo rule, we have 1−cosγ1+cosγ=a+b−ca+b+cwhere a+c=2b ⇒tan2γ2=a+b−c3b ∴tan2α2+tan2β2=b+c−a+a+b−c3b=2b3b=23