In a triangle ABC, if c=(a-b)secθ, then tanθ is equal to
2aba-bsinC2
2aba-btanC2
2aba-bsecC2
2aba-bcosC2
Determine the value of tanθ
Given, c=(a-b)secθ
⇒secθ=ca-b...(1)
We know that, tan2θ=sec2θ–1, therefore substituting the value of (1) we have,
⇒tan2θ=ca-b2–1⇒=c2(a-b)2-1⇒=c2-(a-b)2(a-b)2⇒=(c-a+b)×(c+a-b)(a-b)2⇒=(2s-2a)(2s-2b)(a-b)2[∴s=a+b+c2]⇒=4×(s-a)(s-b)(a-b)2⇒tanθ=2×(s-a)(s-b)(a-b)⇒tanθ=2×aba-bsinC2[∵sinC2=(s-a)(s-b)ab]
Hence, option (A) is correct.