sec16cosec74-cot74tan16
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4
sec(90-16)cosec74 - cot74tan(90-16)
because sec(90 - A)= cosecA and tan(90 - A)= cotA
we get
cosec74cosec74-cot74cot74
cosec^2*74 - cot^3*74
this can be observed as an identity
cosec^2*a - cot^2*a= 1
so we get the final answer as
1
because sec(90 - A)= cosecA and tan(90 - A)= cotA
we get
cosec74cosec74-cot74cot74
cosec^2*74 - cot^3*74
this can be observed as an identity
cosec^2*a - cot^2*a= 1
so we get the final answer as
1
Anonymous:
should not that be sec(90-74)
.?
oops sorry
its okwy
thanks ill correct it
unfortunately i cant
no worries :-)
Answered by
0
Step-by-step explanation:
sec(90-74) x cosec74-cot74 x tan(90-74)
As we can write sec(90-a) as Coseca
cosec74xcosec74- cot74 x cot74
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