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If `tanbeta = costheta*tanalpha` , then `tan^2(theta/2)` is equal toA. `(sin(alpha+beta))/(sin(alpha-beta))`B. `(cos(alpha-beta))/(cos(alpha+beta))`C. `(sin(alpha-beta))/(sin(alpha+beta))`D. `(cos(alpha+beta))/(cos(alpha-beta))` |
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Answer» Correct Answer - C `"tan"^(2)(theta)/(2)=(1-cos theta)/(1+cos theta)=(tan alpha - tan beta)/(tan alpha + tan beta)=(sin(alpha-beta))/(sin(alpha+beta))` |
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