Question

Prove that a square + b square minus c square by c square + a square minus b square is equal to tan b by tanc​

Answer 1

$\underline{\textsf{To prove:}}$

$\textsf{In triangle ABC}$

$\mathsf{\dfrac{a^2+b^2-c^2}{c^2+a^2-b^2}=\dfrac{tanB}{tanC}}$

$\underline{\textsf{Solution:}}$

$\underline{\mathsf{Cosine\;formula:}}$

$\mathsf{In\;\triangle\;ABC}$

$\mathsf{cosA=\dfrac{b^2+c^2-a^2}{2bc}}$

$\mathsf{cosB=\dfrac{c^2+a^2-b^2}{2ca}}$

$\mathsf{cosC=\dfrac{a^2+b^2-c^2}{2ab}}$

$\underline{\mathsf{Sine\;formula}}$

$\mathsf{In\;\traingle\;ABC}$

$\mathsf{\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}=2R}$

$\mathsf{Consider}$

$\mathsf{\dfrac{a^2+b^2-c^2}{c^2+a^2-b^2}}$

$\textsf{Using Cosine formula,}$

$\mathsf{=\dfrac{2ab\;cosC}{2ca\;cosB}}$

$\mathsf{=\dfrac{b\;cosC}{c\;cosB}}$

$\textsf{Using Sine formula,}$

$\mathsf{=\dfrac{2R\,sinB\;cosC}{2R\,sinC\;cosB}}$

$\mathsf{=\dfrac{sinB\;cosC}{sinC\;cosB}}$

$\mathsf{=\dfrac{sinB}{cosB}{\times}\dfrac{cosC}{sinC}}$

$\mathsf{=\dfrac{\dfrac{sinB}{cosB}}{\dfrac{sinC}{osC}}}$

$\mathsf{=\dfrac{tanB}{tanC}}$

$\implies\boxed{\mathsf{\dfrac{a^2+b^2-c^2}{c^2+a^2-b^2}=\dfrac{tanB}{tanC}}}$

$\underline{\textsf{Find more:}}$

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