Question

{cosec²(90-A)-tan²A /2(cos²48+cos²42)} -{2tan²30*sec²52*sin²38/
cosec² 70-tan²20}
evaluate

Answer 1

hope it will help you.

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