हल करें `(1- tan theta)(1+ sin 2 theta)=(1+tan theta)`

Explanation

प्रश्न से `(1-tan theta)(1+(2tan theta)/(1+tan^(2) theta))=1+ tan theta`
या `(1-tan theta) ((1+tan^(2)theta+ 2 tan theta))/(I+ tan^(2) theta)=1+ tan theta`
या `(t-tan theta) (1+tan theta)^(2)-(1+tan theta)(1+tan^(2) theta)`
या `(1+ tan theta)[(1-tan theta)(1+tan theta)-(1+ tan^(2) theta)]=0`
या `(t+tan theta)[1-tan^(2) theta-1-tan^(2)theta]=0`
या `(1+tan theta)(-2 tan^(2) theta)=0`
या `tan^(2) theta(1+ tan theta)=0`
यदि `tan^(2) theta=0`, तो `tan theta=0 rArr 0 n pi`
यदि `1+tan theta=0` तो `tan theta=-1= tan (-(pi)/(4))`
`:. theta=npi+(-(pi)/(4))=npi-(pi)/(4)`
अतः `theta=npi, n pi-(pi)/(4), n in Z`