Advertisements
Advertisements
प्रश्न
Evaluate the following without using a table:
2 tan 18° tan 72° − cot2 30° + cos 34° cosec 56°
बेरीज
Advertisements
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]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
