HOLA BRAINIACS!
Please solve this.......
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secx+tanx=k
[1/cosx+sinx/cosx]=k
(1+sinx)/cosx=k
(1+sinx)=kcosx
(1+sinx)²=k²cos²k
1+2sinx+sin²x=k²(1-sin²x)
1+2sinx+sin²x=k²-k²sin²x
(k²+1)sin²x+2sinx+1-k²=0
use the quadratic formula to solve:
sinx=t
(k²+1)t²+2t+1-k²=0
-2±√(2)²-4(k²+1)(1-k²) / 2k²+2
-2±√4k⁴/ 2k²+2 =>-2±2k² / 2k²+2 =>-1±k² / k²+1 =>
-1-k² / k²+1 or -1+k² / k²+1
In conclusion sinx=t =>sinx= (k²-1)/(k²+1)
[1/cosx+sinx/cosx]=k
(1+sinx)/cosx=k
(1+sinx)=kcosx
(1+sinx)²=k²cos²k
1+2sinx+sin²x=k²(1-sin²x)
1+2sinx+sin²x=k²-k²sin²x
(k²+1)sin²x+2sinx+1-k²=0
use the quadratic formula to solve:
sinx=t
(k²+1)t²+2t+1-k²=0
-2±√(2)²-4(k²+1)(1-k²) / 2k²+2
-2±√4k⁴/ 2k²+2 =>-2±2k² / 2k²+2 =>-1±k² / k²+1 =>
-1-k² / k²+1 or -1+k² / k²+1
In conclusion sinx=t =>sinx= (k²-1)/(k²+1)
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