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Question Number 132081 by liberty last updated on 11/Feb/21
Prove tan (A+B)−tan A= ((sin B)/(cos A cos (A+B)))
$$\mathrm{Prove}\:\mathrm{tan}\:\left(\mathrm{A}+\mathrm{B}\right)−\mathrm{tan}\:\mathrm{A}=\:\frac{\mathrm{sin}\:\mathrm{B}}{\mathrm{cos}\:\mathrm{A}\:\mathrm{cos}\:\left(\mathrm{A}+\mathrm{B}\right)} \\ $$
Answered by rs4089 last updated on 11/Feb/21
tan(A+B)−tanA  ((sin(A+B))/(cos(A+B)))−((sinA)/(cosA))  ((sin(A+B).cosA−sinA.cos(A+B))/(cos(A+B).cosA))  ((sin{(A+B)−(A)})/(cos(A+B).cosA))  ((sinB)/(cos(A+B).cosA))
$${tan}\left({A}+{B}\right)−{tanA} \\ $$$$\frac{{sin}\left({A}+{B}\right)}{{cos}\left({A}+{B}\right)}−\frac{{sinA}}{{cosA}} \\ $$$$\frac{{sin}\left({A}+{B}\right).{cosA}−{sinA}.{cos}\left({A}+{B}\right)}{{cos}\left({A}+{B}\right).{cosA}} \\ $$$$\frac{{sin}\left\{\left({A}+{B}\right)−\left({A}\right)\right\}}{{cos}\left({A}+{B}\right).{cosA}} \\ $$$$\frac{{sinB}}{{cos}\left({A}+{B}\right).{cosA}} \\ $$

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