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Simplify`[(3sin2alpha)/(5+3cos2alpha)]+tan^(-1)[(tanalpha)/4],w h e r e-pi/2 |
Answer» `tan^(-1)[(3sin2alpha)/(5+3cos2alpha)]+tan^(-1)[tanalpha/4]` `tan^(-1)[(6tanalpha)/(8+2tan^2alpha)]+tan^(-1)(tanalpha/4)` `(6tan^2alpha)/(4(8+2tan^2alpha))<1` `tan^(-1)[((3tanalpha)/(4ttan^2alpha)+tanalpha/4)/(1-(3tanalpha)/(4+tan^2alpha)*tanalpha/4)]` `tan^(-1)((12tanalpha+4tanalpha+tan^3alpha)/(16+tan^2alpha))` `tan^(-1)((16tanalpha+tan^3alpha)/(16+tan^2alpha))` `tan^(-1)((tanalpha(16+tan^2alpha))/(16tan^2alpha))` `tan^(-1)tanalpha=alpha`. |
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