If $$sin$$θ$$cos3$$θ$$+sin3$$θ$$cos9$$θ$$+sin9$$θ$$cos27$$θ$$=AtanB$$θ−tanCθ, then $$B+CA=$$ A$$0$$,θ≠$$2n+1$$$$π$$$$2$$,n∈Z

Question

If  $$sin$$θ$$cos3$$θ$$+sin3$$θ$$cos9$$θ$$+sin9$$θ$$cos27$$θ$$=AtanB$$θ−tanCθ, then $$B+CA=$$                             A>$$0$$,θ≠$$2n+1$$$$π$$$$2$$,n∈Z

Infinity Learn · Accepted Answer

We  have  $$sin$$θ$$cos3$$θ$$=12sin2$$θ$$cos$$θ$$cos3$$θ $$12sin(3$$θ$$-$$θ$$)$$$$cos$$θ$$cos3$$θ$$=12sin3$$θ$$cos$$θ$$-cos3$$θ$$sin$$θ$$cos$$θ$$cos3$$θ  $$sin$$θ$$cos3$$θ$$=12tan3$$θ−$$tan$$θ   $$-----(1)Similarly$$   $$sin3$$θ$$cos9$$θ$$=12tan9$$θ−$$tan3$$θ $$----(2)and$$           $$sin9$$θ$$cos27$$θ$$=12tan27$$θ−$$tan9$$θ  $$----(3)adding$$ (1) ,(2) ,(3)∴   $$sin$$θ$$cos3$$θ$$+sin3$$θ$$cos9$$θ$$+sin9$$θ$$cos27$$θ$$=12tan27$$θ−$$tan$$θ$$=AtanB$$θ−tanCθ∴$$B+CA=27+112=56$$

If $\frac{\sin θ}{\cos 3 θ} + \frac{\sin 3 θ}{\cos 9 θ} + \frac{\sin 9 θ}{\cos 27 θ} = A (\tan B θ - \tan C θ)$ , then $(\frac{B + C}{A}) = (A > 0, θ \neq (2 n + 1) \frac{π}{2}, n \in Z)$

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If sinθcos3θ+sin3θcos9θ+sin9θcos27θ=AtanBθ−tanCθ, then B+CA= A>0,θ≠2n+1π2,n∈Z

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If $\frac{\sin θ}{\cos 3 θ} + \frac{\sin 3 θ}{\cos 9 θ} + \frac{\sin 9 θ}{\cos 27 θ} = A (\tan B θ - \tan C θ)$ , then $(\frac{B + C}{A}) = (A > 0, θ \neq (2 n + 1) \frac{π}{2}, n \in Z)$