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jeyben [28]
4 years ago
7

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

Mathematics
2 answers:
nekit [7.7K]4 years ago
7 0
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.
Julli [10]4 years ago
5 0
   
\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)= \\  \\ 
= \frac{1}{\cos^2(53^o)} - \tan^2 (53^o)=\\\\
= \frac{\sin^2(53^o)+\cos^2(53^o)}{\cos^2(53^o)} - \tan^2 (53^o)=


\displaystyle\\
= \frac{\sin^2(53^o)}{\cos^2(53^o)} +\frac{\cos^2(53^o)}{\cos^2(53^o)} - \tan^2 (53^o)=  \\  \\ 
= \frac{\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}}



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