Math, asked by Paracetamolstuda, 1 year ago

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Without using trigonometrical tables, evaluate:

(5 cos 0° - 2 sin 30° + √3 cos 30°)/
tan 30" x tan 60° x cos 60°++ 3 sin 29° sec 61°​

Answers

Answered by abhi178
48

(5cos0° - 2sin30° + √3cos30°)/(tan30° × tan60° × cos60° + 3sin29° sec61°) = 11/7

we know, sin30° = 1/2 , cos30° = √3/2

cos0° = 1 , cos60° = 1/2,

tan30° = 1/√3 , tan60° = √3

so,

= (5 × 1 - 2 × 1/2 + √3 × √3/2)/(1/√3 × √3 × 1/2 + 3sin29° sec61°)

= (5 - 1 + 3/2)/[1/2 + 3sin(90° - 61°) sec61°]

we know, sin(90° - θ) = cosθ

so, sin(90° - 61°) = cos61°

also secθ = 1/cosθ

= (4 + 3/2)/[1/2 + 3cos61° × 1/cos61°]

= [(8 + 3)/2]/[1/2 + 3]

= (11/2)/(7/2)

= 11/7

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