Question

prove that (1+tan17°) (1+tan28°) =2

Answer 1

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

$\mathsf{(1+tan17^{\circ})(1+tan28^{\circ})=2}$

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

$\mathsf{Consider,}$

$\mathsf{17^{\circ}+28^{\circ}=45^{\circ}}$

$\mathsf{tan(17^{\circ}+28^{\circ})=tan45^{\circ}}$

$\mathsf{Using,\boxed{\bf\;tan(A+B)=\dfrac{tanA+tanB}{1-tanA\,tanB}}}$

$\implies\mathsf{\dfrac{tan17^{\circ}+tan28^{\circ}}{1-tan17^{\circ}\,tan28^{\circ}}=1}$

$\implies\mathsf{tan17^{\circ}+tan28^{\circ}=1-tan17^{\circ}\,tan28^{\circ}}$

$\implies\mathsf{tan17^{\circ}+tan28^{\circ}+tan17^{\circ}\,tan28^{\circ}=1}$

$\textsf{Add 1 on bothsides, we get}$

$\implies\mathsf{1+tan17^{\circ}+tan28^{\circ}+tan17^{\circ}\,tan28^{\circ}=2}$

$\implies\mathsf{1(1+tan17^{\circ})+tan28^{\circ}(1+tan17^{\circ})=2}$

$\therefore\boxed{\mathsf{(1+tan17^{\circ})(1+tan28^{\circ})=2}}$

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

यदि A+B+C= 90 तो सिद्ध कीजिए: tanAtanB+tanBtanC+tanCtanA=1

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