If `sinx+sin^2x=1` then the value of `tan^8x-tan^4x-2tan^2x+1` will be equal to 0 (b) 1 (c) 2 (d) 3
If `sinx+sin^2x=1` then the value of `tan^8x-tan^4x-2tan^2x+1` will be equal to 0 (b) 1 (c) 2 (d) 3
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Correct Answer - C
`secx=1-sec^2x`
`rArrsecx=-tan^2x`
Squaring both side, we get
`sec^2x=tan^4x`
`rArr1+tan^2x=tan^4x`
Squaring both side, we get
`tan^8x=tan^4+2tan^2x+1`
`rArr tan^8x-tan^4x-2tan^2x=1`
`tan^8x-tan^4x-2tan^2x+1=2`
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