Question

Evaluate 2(cos 58\sin32) - root 3(cos 38 cos 52\tan15 tan60 tan 75)

Answer 1

Answer:

Step-by-step explanation:

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Step-by-step explanation:

2(cos58/sin32)-√3(cos38*cosec52/tan15*tan60*tan75)

= 2{cos58/sin(90-58)}-√3[{cos38*cosec(90-38)}/tan(90-75)*tan60*tan75]

= 2{cos58/cos58}-√3[cos38*sec38/cot75*tan60*tan75]                  (1/tanФ = cotФ)

= 2*1-√3[1/tan60]                                               

= 2-√3(1/√3)

= 2-1

= 1

Answer 2

2(cos 58\sin32) - root 3(cos 38 cos 52\tan15 tan60 tan 75) = 1

Step-by-step explanation:

2($\frac{cos 58}{sin32}$) - $\sqrt{3}$($\frac{cos 38 cos 52}{tan15 tan60 tan 75}$)

= 2($\frac{sin 90 - 58}{sin 32}$) - $\sqrt{3}$($\frac{sin (90 - 38) cosec 52}{tan15 tan60 tan 75}$)             (∵ sin (90 - θ) = cosθ)

= 2($\frac{sin 32}{sin 32}$) - $\sqrt{3}$($\frac{sin 52 cosec 52}{cot(90 - 15) tan60 tan 75}$}              (∵ cot(90−θ) = tanθ)

= 2($\frac{sin 32}{sin 32}$) - $\sqrt{3}$($\frac{sin 52 cosec 52}{cot 75 tan60 tan 75 }$)

= 2  - $\sqrt{3}$(1/tan 60)                    ( ∵sinθ cosec θ = 1 and cotθ tanθ = 1)

= 2 - $\sqrt{3}$ (1/ $\sqrt{3}$)                        (∵ tan60 = $\sqrt{3}$)

= 2 - 1

= 1

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