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महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Prove the following: tan6° tan42° tan66° tan78°= 1

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प्रश्न

Prove the following:

tan6° tan42° tan66° tan78° = 1

बेरीज
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उत्तर

L.H.S. = tan6° tan42° tan66° tan78°

= `sin6^circ/cos6^circ  * sin42^circ/cos42^circ  *sin66^circ/cos66^circ  *sin78^circ/cos78^circ`

= `((2sin66^circ sin6^circ)(2sin78^circ sin42^circ))/((2cos66^circ cos6^circ)(2cos78^circ cos42^circ)`

= `(cos(66^circ - 6^circ) - cos(66^circ + 6^circ))/(cos(66^circ + 6^circ) + cos(66^circ - 6^circ))  * (cos(78^circ - 42^circ) - cos(78^circ + 42^circ))/(cos(78^circ + 42^circ) + cos(78^circ - 42^circ))`

= `((cos60^circ -  cos72^circ)(cos36^circ -  cos120^circ))/((cos60^circ +  cos72^circ)(cos36^circ +  cos120^circ)`

= `((cos60^circ -  sin18^circ)(cos36^circ +  sin30^circ))/((cos60^circ +  sin18^circ)(cos36^circ -  sin30^circ)` ...[∵ cos(90° + θ) = – sin θ]

= `((1/2 - (sqrt(5) - 1)/4)((sqrt(5) + 1)/4 + 1/2))/((1/2 + (sqrt(5) - 1)/4)((sqrt(5) + 1)/4 - 1/2))`

= `(9 - 5)/(5 - 1)`

= 1

= R.H.S.

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पाठ 3: Trigonometry - 2 - Miscellaneous Exercise 3 [पृष्ठ ५८]

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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. (28) | पृष्ठ ५८

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