1)Show that tan^2x-tan^2xsin^2x+cos^2x=1.
NEED STEP, PLZ!!!
Maths problem
2014-06-07 6:24 am
回答 (3)
2014-06-07 6:45 am
✔ 最佳答案
tan²x - tan²x sin²x + cos²x= tan²x (1 - sin²x) + cos²x
= tan²x cos²x + cos²x
= ( tan²x + 1 ) cos²x
= ( sin²x/cos²x + cos²x/cos²x ) cos²x
= [ (sin²x + cos²x) / cos²x ] cos²x
= (1 / cos²x) cos²x
= 1
2014-06-07 5:45 pm
tan²x - tan²x sin²x + cos²x
= tan²x (1 - sin²x) + cos²x
= tan²x cos²x + cos²x
= sin²x/cos²x * cos²x + cos²x
= sin²x + cos²x
= 1
= tan²x (1 - sin²x) + cos²x
= tan²x cos²x + cos²x
= sin²x/cos²x * cos²x + cos²x
= sin²x + cos²x
= 1
2014-06-07 6:49 am
Show that (tan x)^2 - (tan x)^2 (sin x)^2 + (cos x)^2 = 1
L.H.S.=(sin x)^2/(cos x)^2 - (sin x)^4/(cos x)^2 + (cos x)^4/(cos x)^2
={ (sin x)^2 - [(sin x)^2]^2 - [(cos x)^2]^2] } / (cos x)^2 Note: (a+b)(a-b)=a^2 - b^2
={ (sin x)^2 - [(sin x)^2 + (cos x)^2 ][ (sin x)^2 - (cos x)^2] } / (cos x)^2
Note: (sin x)^2 + (cos x)^2 = 1
={ (sin x)^2 - [(sin x)^2 - (cos x)^2] } / (cos x)^2
=(cos x)^2 / (cos x)^2
=1
=R.H.S.
L.H.S.=(sin x)^2/(cos x)^2 - (sin x)^4/(cos x)^2 + (cos x)^4/(cos x)^2
={ (sin x)^2 - [(sin x)^2]^2 - [(cos x)^2]^2] } / (cos x)^2 Note: (a+b)(a-b)=a^2 - b^2
={ (sin x)^2 - [(sin x)^2 + (cos x)^2 ][ (sin x)^2 - (cos x)^2] } / (cos x)^2
Note: (sin x)^2 + (cos x)^2 = 1
={ (sin x)^2 - [(sin x)^2 - (cos x)^2] } / (cos x)^2
=(cos x)^2 / (cos x)^2
=1
=R.H.S.
參考: ME
收錄日期: 2021-04-15 15:43:07
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