If `sintheta+ costheta=1`, then find the general value of `theta`.

Options

A. `n pi`
B. `(npi)/2`
C. `0`
D. none of these

Answer

B

Explanation

Given that, `sintheta+costheta=1`
On squaring both sides, we get ltBrgt `" "sin^(2)theta+cos^(2)theta+2sintheta*costheta=1` ltBrgt `rArr" "1+2sintheta*cos=1" "[becausesin2x=2sinxcosx]`
`rArr" "sin2theta=0rArr2theta=npi+(-1)^(n)*0`
`therefore" "theta=(npi)/(2)`
Alternate Method
` " "sintheta+costheta=1`
`rArr" "(1)/(sqrt( 2))*sintheta +(1)/(sqrt(2))*costheta=(1)/(sqrt(2))`
`rArr" "sintheta*cos""(pi)/(4)+costheta*sin""(pi)/(4)=(1)/(sqrt(2))" "[becausesin""(pi)/(4)=(1)/(sqrt(2))=cos""(pi)/(4)]`
`rArr" "sin(theta+(pi)/(4)) =sin""(pi)/(4)" " [ because sin(x+y)=sinx*cosy+cosx*siny]`
`rArr" "theta+(pi)/(4)=npi+(-1)^(n)(pi)/(4)`
`therefore" "theta=npi+(-1)^(n)(pi)/(4)-(pi)/(4)`