1.

`sec^(2)2x=1-tan2x`

Answer» Given equation is `sec^(2)2x=1-tan2x`
or `1+tan^(2)2x=1-tan2x`
or `tan^(2)2x+tan2x=0`
or `tan2x(1+tan2x)=0`
`rArr tan2x=0` or `1+tan2x=0`
If `tan2x=0`, then `2x=npi`
`rArr x=(npi)/(2)` where `n in Z`
and `tan2x = 1=tan(pi-pi/4)`
`rArr tan2x=tan(3pi)/(4)`
`rArr 2x=npi + (3pi)/(4)`
`rArr x=(npi)/(2) + (3pi)/(8)` where `n in Z`
Therefore, the general solution of given equations is
`x=(npi)/(2)` or `(npi)/(2)+(3pi)/(8), n in Z` Ans.


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