Question

यदि A+B+C= 900 तो सिद्ध कीजिए: tanAtanB+tanBtanC+tanCtanA=1

Answer 1

$\textbf{Given:}$

$A+B+C=90^{\circ}$

$\implies\;A+B=90^{\circ}-C$

$\implies\;tan(A+B)=tan(90^{\circ}-C)$

$\implies\;tan(A+B)=cotC$

$\text{Using, }\boxed{\bf\;tan(A+B)=\frac{tanA+tanB}{1-tanA\;tanB}\;\text{and}\;cot\theta=\frac{1}{tan\theta}}$

$\implies\frac{tanA+tanB}{1-tanA\;tanB}=\frac{1}{tanC}$

$\implies\;tanC(tanA+tanB)=1-tanA\;tanB$

$\implies\;tanC\;tanA+tanB\;tanC=1-tanA\;tanB$

$\implies\boxed{\bf\;tanA\;tanB+tanB\;tanC+tanC\;tanA=1}$

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Answer 2

Answer:

A+B+C=90∘

\implies\;A+B=90^{\circ}-C⟹A+B=90∘−C

\implies\;tan(A+B)=tan(90^{\circ}-C)⟹tan(A+B)=tan(90∘−

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