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58° o sin 22° cos 38° cosec 52°32° .cos68° \/_3 (tan 18° tan 35° tan 60° tan 72° tan |
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Answer» Question is incomplete please post the full question Assuming that you meancos 58/sin 32 + sin 22/cos 68- (cos 38 csc 52)/(tan 18*tan 35*tan 60*tan 72*tan 55) Use the cofunction identities (in degrees):sin(90 - t) = cos t, cos(90 - t) = sin t, and tan(90 - t) = cot t. This simplifies the expression in question tosin(90 - 58)/sin 32 + cos(90 - 22)/cos 68- (cos 38 csc 52)/(tan 18*tan 35*tan 60*cot(90-72)*cot(90-55)) = 1 + 1 - (cos 38 csc 52)/(tan 18*tan 35*tan 60*cot 18*cot 35) Now, use csc t = 1/sin t and cot t = 1/tan t:2 - (cos 38/sin 52)/(tan 18*tan 35*tan 60*(1/tan 18)*(1/tan 35))= 2 - (cos 38/sin 52)/tan 60= 2 - (cos 38/cos(90-52))/tan 60, via cofunction identity= 2 - 1/tan 60= 2 - 1/√3. Like my answer if you find it useful! |
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