Prove that of the two trigonometric equations. tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B sin⁡2θ=2sin⁡θcos⁡θ

Question

Prove that of the two trigonometric equations.  tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B  sin⁡2θ=2sin⁡θcos⁡θ

brawnyN · Accepted Answer

Solution:  tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B  Now first we take LHS of this:  tan⁡(A+B)=sin⁡(A+B)cos⁡(A+B)  we know sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B  and cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B  Now put in eauqtion:  tan⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡Bcos⁡Acos⁡B−sin⁡Asin⁡B  devide numerator cos⁡Acos⁡B: and also devide denomianator.  tan⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡Bcos⁡A⋅cos⁡Bcos⁡Acos⁡B−sin⁡A⋅sin⁡Bcos⁡A⋅cos⁡B  tan⁡(A+B)=sin⁡Acos⁡Bcos⁡Acos⁡B+cos⁡Asin⁡Bcos⁡Acos⁡Bcos⁡Acos⁡Bcos⁡Acos⁡B−sin⁡Asin⁡Bcos⁡Acos⁡B  tan⁡(A+B)=sin⁡Acos⁡A+sin⁡Bcos⁡B1−sin⁡Acos⁡A⋅sin⁡Bcos⁡B  tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B = RHS  Solution:  sin⁡2θ=2sin⁡θcos⁡θ  Now take LHS of this:

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Prove that of the two trigonometric equations. \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}

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