$a \tan θ + b s e c θ = c$ সমীকরণের মূলদ্বয় $α, β$ হলে, প্রমাণ কর যে, $\tan (α + β) = \frac{2 c a}{a^{2} - c^{2}}$ ।

Created: 1 year ago | Updated: 5 months ago

Updated: 5 months ago

$a \tan θ + b s e c θ = c \Rightarrow a \tan θ - c = - b s e c θ$

$a^{2} \tan^{2} θ + c^{2} - 2 c a \tan θ = b^{2} + b^{2} \tan^{2} θ$

$\Rightarrow (a^{2} - b^{2}) \tan^{2} θ - 2 c a \tan θ + c^{2} - b^{2} = 0$

$\tan α + \tan β = \frac{2 c a}{a^{2} - b^{2}}; \tan α \tan β = \frac{c^{2} - b^{2}}{a^{2} - b^{2}}$

$L . H . S = \tan (α + β) = \frac{\tan α + \tan β}{1 - \tan α \tan β} = \frac{\frac{2 c a}{a^{2} - b^{2}}}{1 - \frac{c^{2} - b^{2}}{a^{2} - b^{2}}}$

$= \frac{2 c a}{a^{2} - b^{2} - c^{2} + b^{2}} = \frac{2 c a}{a^{2} - c^{2}} = R . H . S (P r o v e d)$

1 year ago

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