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