Question

Help me solve this question

Answer 1

Consider the first term on the left side.

`\displaystyle (2sin(A))/(cos(3A))\\\\=(2sin(A)cos(A))/(cos(3A)cos(A))\qquad\text{multiply by cos(A)/cos(A)}\\\\=(sin(2A))/(cos(3A)cos(A))\qquad\text{double angle formula for sin}\\\\=(sin((2A-A)))/(cos(3A)cos(A))=(sin(3A)cos(A)-cos(3A)sin(A))/(cos(3A)cos(A))\\\\=tan(3A)-tan(A)`

Then the sum of the three terms on the left is

`\left(tan(3A)-tan(A)\right)+\left(tan(9A)-tan(3A)\right)+\left(tan(27A)-tan(9A)\right)\\\\=tan(27A)-tan(A)\qquad\text{Q.E.D.}`