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Evaluate the following without using a table: 2 tan 18° tan 72° − cot2 30° + cos 34° cosec 56° - Mathematics

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प्रश्न

Evaluate the following without using a table:

2 tan 18° tan 72° − cot2 30° + cos 34° cosec 56°

बेरीज
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उत्तर

Given:

2 tan 18° tan 72° − cot2 30° + cos 34° cosec 56°

Step 1: 

tan A ⋅ tan(90° − A) = 1 ⇒ tan 18° ⋅ tan 72° = 1

2 ⋅ tan 18° ⋅ tan 72° = 2 ⋅ 1 = 2

Step 2: 

cot 30° = `1/(tan 30°)`

= `(1/1)/sqrt3 = sqrt3 ⇒ cot^2 30° = (sqrt3)^2 = 3`

Step 3: 

cos34° ⋅ csc56°

cosec 56° = `1/(sin 56°) and sin 56° = cos(34°)`

cosec 56° = `1/(cos 34°) ⇒ cos 34° ⋅  "cosec"  56° = cos 34° ⋅ 1/cos 34° = 1`

2 tan 18° tan 72° − cot2 30° + cos 34° cosec 56°

= 2 − 3 + 1 = 0     ... [Putting all values together]

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पाठ 19: Trigonometry - MISCELLANEOUS EXERCISE [पृष्ठ २३९]

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बी निर्मला शास्त्री Mathematics [English] Class 9 ICSE
पाठ 19 Trigonometry
MISCELLANEOUS EXERCISE | Q IV. 9. | पृष्ठ २३९
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