Question

cos 58/sin 22+sin22/cos68-cos 38 cosec 52/tan 18 tan 35 tan 60 tan 72 tan 55,Evaluate it.

Answer 1

I think question is ---->
$\bf{\frac{cos68}{sin22}+\frac{sin22}{cos68}-\frac{cos38.cosec52}{tan18.tan35.tan60.tan72.tan55}}$

we know, sin(90- x) = cosx, cos(90- x) = sinx 
tan(90- x) = cotx , cot(90- x) = tanx 
cosec(90 - x) = secx , sec(90 - x) = cosecx

so, cos68° = cos(90° - 22°) = sin22° ---(1)
cos38° = cos(90° - 52°) = sin52° ----(2)
tan18° = tan(90° - 72°) = cot72° ------(3)
tan35° = tan(90° -55°) = cot55° -----(4)

from (1), (2), (3) and (4), 

= sin22°/sin22° + sin22°/sin22° - (sin52°.cosec52°)/(cot72°.cot55°.tan60°.tan72°.tan55°)

= 1 + 1 - (sin52° × 1/sin52°)/{(tan72° × cot72°) × (tan55°.cot55°) × tan60°}

= 1 + 1 - 1/(1 × 1 × tan60° )

= 2 - 1/√3 [ because tan60° = √3]

= 2 - √3/3 = (6 - √3)/3

Answer 2

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