evaluate the following
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use sin x = cos(90-x)
cos x = sin(90-x)
tan x = cot(90-x)
tan 5 = cot(90-5)=cot85
tan 25= cot(90-25) = cot65
cos x = sin(90-x)
tan x = cot(90-x)
tan 5 = cot(90-5)=cot85
tan 25= cot(90-25) = cot65
AmanRajSinghania:
of which class? 9 or 10
10
ohh
now you understood?
not exactly wat to do with all the values of cot now
cot85 x tan85 = 1 and cot 65 x tan65 =1
and put tan45 = 1..so denominator becomes 1
ohk i seem to het it now
thanku so much
wlcm!!
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