Question
Download Solution PDFtan2α =
[√3 = 1.732 प्रयोग करें]
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है:
tan2α =
प्रयुक्त सूत्र:
Tan15° = Tan(45 – 30)°
त्रिकोणमिति सूत्र से, हम जानते हैं,
Tan (A – B) = (Tan A – Tan B) /(1 + Tan A Tan B)
गणना:
इसलिए, हम लिख सकते हैं,
tan(45 – 30)° = tan 45° – tan 30°/1+ tan 45° tan 30°
अब tan 45° और tan 30° का मान तालिका से रखने पर, हमें प्राप्त होता है;
tan(45 – 30)° = (1 – 1/√3)/ (1 + 1.1/√3)
tan (15°) = √3 – 1/ √3 + 1
= (√3 – 1)2/ [(√3)2 - 12]
= (3 + 1 - 2√3)/2 = 2 - √3
= 0.268
अत:, tan (15°) का मान 0.268 है।
Alternate Method
|
tan 30° = tan 2(15°)
त्रिकोणमिति सूत्र से हम जानते हैं,
tan2α =
गणना
इसलिए, हम लिख सकते हैं,
tan 30° = 2 × tan 15° /(1 - tan 2 15°)
अब tan 30° का मान रखने पर हमें प्राप्त होता है;
⇒ 1/ √3 = 2 tan 15° / (1 - tan 2 15°)
माना tan (15°) = x
⇒ 1/ √3 = 2x / (1 - x 2 )
⇒ x 2 - 1 + 2√3 x = 0
⇒ x2 + 2√3 x - 1 = 0
द्विघात सूत्र से,
x =
⇒ x =
⇒ x =
⇒ (4 - 2√3)/2 = 2 - √3
= 0.268
अत:, tan (15°) का मान 0.268 है।
Last updated on Jul 12, 2025
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