Question

Evaluate cos 58 sin 32 sin 22 cos 68 â cos 38 cos 52 tan18 tan35 tan 60 tan 72 tan 55

Answer 1

Answer:

Step-by-step explanation

⇒(cos(90-32)/sin32)+(sin(90-68)/cos68)-cos(90-52)cosec52)/tan(90-72).tan35. tan60.tan72.tan(90-35)

⇒(sin32/sin32)+(cos68/cos68)-sin52.cosec52/cot72.tan35.tan60.tan72.cot35

⇒1+1-1/1*1*tan60

⇒2-1/√3

⇒(2√3-1)/√3

by multiply nominator and denominator by √3

⇒(6-√3)/3

Answer 2

Answer:

Step-by-step explanation:

PLEASE MARK ME AS BRAINLIEST  

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