`(tan alpha + tan beta )/( tan (alpha + beta ))+(tan alpha - tan beta )/(tan (alpha - beta ))` =

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If 2tan alpha=3tan beta then tan(alpha-beta)=

(1+tan alpha tan beta)^2 + (tan alpha - tan beta)^2 =

(1+tan alpha tan beta)^(2) + (tan alpha -tan beta)^(2) =

(1+tan alpha tan beta)^(2)+(tan alpha-tan beta)^(2)=

Prove that (tan alpha+tan beta)/(cot alpha+cot beta)=tan alpha tan beta

sin alpha + sin beta = a "and" cos alpha + cos beta = b (tan alpha + tan beta)/(1-tan alpha tan beta)=

(no alpha + no beta) / (no alpha-no beta) = (tan ((alpha + beta) / (2))) / (tan ((alpha-beta) / (2)))

(sec alpha * sec beta + tan alpha tan beta )^(2)-( sec alpha * tan beta + tan alpha * sec beta )^(2) =