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

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

A

`x=(npi)/(2)+(3pi)/(4),ninZ`

B

`x=(npi)/(3),ninZ`

C

`x=(npi)/(2)+(3pi)/(8),ninZ`

D

None of the above

Text Solution

Verified by Experts

The correct Answer is:
C
`sec^(2)2x=1-tan2x`
`1+tan^(2)2x=1-tan2x["becausesec^(2)x-tan^(2)x=1]`
`rArrtan2x(tan2x+1)=0`
`rArrtan2x=0ortan2x=-1`
when tan 2x=0
Then , `2x=npirArrx=(npi)/(2)`
When , tan 2x=-1
Then `tan2x="tan"(pi)/(4)`
`tan2x=tan(pi-(pi)/(4))="tan"(3pi)/(4)`
`tan2x=tan(pi-(pi)/(4))="tan"(3pi)/(4)`
`therefore2x=npi+(3pi)/(4)rArrx=(npi)/(2)+(3pi)/(8),ninZ`
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