Question

FInd exact value of: csc (37) sec(53) - tan (53) cot (37)

Answer 1

`\displaystyle\\ \alpha = 37^o \\ \beta = 53^o \\ \alpha+\beta=37^o+53^o=90^o \\ \Longrightarrow~~\alpha \text{ and } \beta \text{ are complementary.} \\ \\ \csc (37^o) \sec(53^o) - \tan (53^o) \cot (37^o)= \\ \\ =\csc (90^o-53^o) \sec(53^o) - \tan (53^o) \cot (90-53^o)= \\ \\ =\sec (53^o) \sec(53^o) - \tan (53^o) \tan (53^o)= \\ \\ =\sec^2 (53^o) - \tan^2 (53^o)= \\ \\ = (1)/(\cos^2(53^o)) - \tan^2 (53^o)=\\\\ = (\sin^2(53^o)+\cos^2(53^o))/(\cos^2(53^o)) - \tan^2 (53^o)=`



`\displaystyle\\ = (\sin^2(53^o))/(\cos^2(53^o)) +(\cos^2(53^o))/(\cos^2(53^o)) - \tan^2 (53^o)= \\ \\ = (\sin^2(53^o))/(\cos^2(53^o)) +1- \tan^2 (53^o)= \\ \\ =\underline{\tan^2 (53^o)}+1- \underline{\tan^2 (53^o)}= \boxed{\bold{1}}`

Answer 2

csc (37) sec (53) - tan (53) cot (37)
1/sin (37) * 1/cos (53) - sin (53) / cos (53) * cos (37) / sin (37) = (1 - sin (53) cos (37)) / cos (53) sin (37) = (1 - sin (53) sin (90 - 37)) / cos (53) cos (90 - 37) = (1 - sin^2 (53)) / cos^2 (53) = cos^2 (53) / cos^2 (53) = 1.