<Dl> <Dd> tan ⁡ γ 2 = cos ⁡ β tan ⁡ γ 1 tan ⁡ γ 4 = cos ⁡ β tan ⁡ γ 3 (\ displaystyle \ tan \ gamma _ (2) = \ cos \ beta \, \ tan \ gamma _ (1) \ qquad \ tan \ gamma _ (4) = \ cos \ beta \, \ tan \ gamma _ (3)) </Dd> </Dl> <Dd> tan ⁡ γ 2 = cos ⁡ β tan ⁡ γ 1 tan ⁡ γ 4 = cos ⁡ β tan ⁡ γ 3 (\ displaystyle \ tan \ gamma _ (2) = \ cos \ beta \, \ tan \ gamma _ (1) \ qquad \ tan \ gamma _ (4) = \ cos \ beta \, \ tan \ gamma _ (3)) </Dd> <P> If the second universal joint is rotated 90 degrees with respect to the first, then γ 3 = γ 2 + π / 2 (\ displaystyle \ gamma _ (3) = \ gamma _ (2) + \ pi / 2). Using the fact that tan ⁡ (γ + π / 2) = 1 / tan ⁡ γ (\ displaystyle \ tan (\ gamma + \ pi / 2) = 1 / \ tan \ gamma) yields: </P> <Dl> <Dd> tan ⁡ γ 4 = cos ⁡ β / tan ⁡ γ 2 = 1 / tan ⁡ γ 1 = tan ⁡ (γ 1 + π / 2) (\ displaystyle \ tan \ gamma _ (4) = \ cos \ beta / \ tan \ gamma _ (2) = 1 / \ tan \ gamma _ (1) = \ tan (\ gamma _ (1) + \ pi / 2) \,) </Dd> </Dl>

What is the function of universal joint in propeller shaft