Question
Download Solution PDFसमीकरण 7x2 - 6x + 1 = 0 के मूल tan α और tan β हैं, जहाँ 2α और 2β एक त्रिभुज के कोण हैं। निम्नलिखित में से कौन सा सही है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFव्याख्या:
दिया गया है:
tanα और tanβ, 7x2 - 6x + 1 = 0 के मूल हैं
इसलिए,
⇒ tanα + tanβ = 6/7...(1)
⇒ tanα.tanβ = 1/7...(2)
⇒ tan(α+β) = \(\frac{tanα + tanβ}{1- tanα . tanβ} = \frac{\frac{6}{7}}{1- \frac{1}{7}} = 1\)
⇒ α +β = 45°
⇒ 2α +2β = 90°
इसका अर्थ है कि त्रिभुज का तीसरा कोण 90° है।
इसलिए, त्रिभुज समकोण है।
चूँकि
⇒tanα - tanβ = \(\sqrt{(tanα + tanβ)^2 - 4tanα .tanβ} = \frac{2\sqrt2}{7}\)
साथ ही
⇒ tan(α -β) = \(\frac{tanα - tanβ}{1+ tanα .tanβ}\) = \(\frac{1}{2\sqrt2}\)
⇒ tan2(α -β) = \(\frac{2tan(α -β)}{1- tan^2(α-β)}\)
= \(\frac{4\sqrt7}{7}\)
इस प्रकार 2α≠2β
∴ विकल्प (c) सही है।
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