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In ΔABC, A + B + C = π show that tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C - Mathematics and Statistics

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प्रश्न

In ΔABC, A + B + C = π show that

tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

योग
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उत्तर

In ΔABC,

A + B + C = π

∴ 2A + 2B + 2C = 2π

∴ 2A + 2B = 2π – 2C

∴ tan (2A + 2B) = tan (2π – 2C)

∴ `(tan2"A" + tan2"B")/(1 - tan2"A"*tan2"B")` = – tan 2C

∴ tan 2A + tan 2B = − tan 2C . (1 − tan 2A . tan 2B)

∴ tan 2A + tan 2B = – tan 2C + tan 2A· tan 2B· tan 2C

∴ tan 2A + tan 2B + tan 2C = tan 2A · tan 2B· tan 2C

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Factorization Formulae - Trigonometric Functions of Angles of a Triangle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometry - 2 - Exercise 3.5 [पृष्ठ ५४]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 11 Maharashtra State Board
अध्याय 3 Trigonometry - 2
Exercise 3.5 | Q 7 | पृष्ठ ५४

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