0.813
tan(22.5)=0.414213562
Tan 42 degrees = 0.9004
Tan(74 degrees) = 3.487414444.....
tan(-60 degrees) = - sqrt(3)
tan 40o ~= 0.8391
0.813
6.25
0.125
The value of tan A is not clear from the question.However, sin A = sqrt[tan^2 A /(tan^2 A + 1)]
To find the value of (\tan(15^\circ) \tan(195^\circ)), we can use the identity (\tan(195^\circ) = \tan(15^\circ + 180^\circ) = \tan(15^\circ)). Thus, (\tan(195^\circ) = \tan(15^\circ)). Consequently, (\tan(15^\circ) \tan(195^\circ) = \tan(15^\circ) \tan(15^\circ) = \tan^2(15^\circ)). The exact value of (\tan^2(15^\circ)) can be computed, but it is important to note that it will yield a positive value.
tan(135) = -tan(180-135) = -tan(45) = -1
tan(22.5)=0.414213562
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= tan (48.323 deg) = 1.1232
Tan 42 degrees = 0.9004
Tan(74 degrees) = 3.487414444.....
tan(61°) = 1.80405 (rounded)