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\frac { 1 - \operatorname { cos } \theta + \operatorname { sin } \theta } { 1 + \operatorname { cos } \theta + \operatorname { sin } \theta } \text { interms of } \operatorname { tan } \frac { \theta } { 2 } |
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Answer» (1-cosx+sinx)/(1+cosx+sinx)=(2sin^2x/2+2sinx/2cosx/2)/(2cos^2x/2+2sinx/2cosx/2)=2sinx/2(sinx/2+cosx/2)/2cosx/2(sinx/2+cosx/2)=tanx/2 |
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