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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Prove the following: tan 20° tan 80° cot50° = 3

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प्रश्न

Prove the following:

tan 20° tan 80° cot 50° = `sqrt(3)`

बेरीज
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उत्तर

L.H.S. = tan 20° tan 80° cot 50°

= tan 20° tan 80° cot (90° − 40°)

= tan 20° tan 80° tan 40°

= tan 20° tan (60° + 20°) tan (60° − 20°)

`=tan20°((tan60°+tan20°)/(1-tan60°tan20°))((tan60°-tan20°)/(1+tan60°tan20°))`

= `tan20°((sqrt3+tan20°)/(1-sqrt3tan20°))((sqrt3-tan20°)/(1+sqrt3tan20°))`

= tan 20° `[((sqrt(3))^2-tan^2 20°)/(1^2-(sqrt3tan20°)^2)]`

= tan 20° `((3-tan^2 20°)/(1-3tan^2 20°))`

= `(3tan20°-tan^3 20°)/(1-3tan^2 20°)`

= tan 3 (20°)
= tan 60°
= `sqrt3`
= R.H.S.

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पाठ 3: Trigonometry - 2 - Miscellaneous Exercise 3 [पृष्ठ ५७]

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बालभारती Mathematics and Statistics (Arts and Science) Part 1 [English] Standard 11 Maharashtra State Board
पाठ 3 Trigonometry - 2
Miscellaneous Exercise 3 | Q II. (1) | पृष्ठ ५७

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