Question

If sec A = 5/4 , prove that ; 3 sin A - 4 sin³A / 4 cos^​3A - 3 cos A = 3tan A - tan³ A/ 1-3 tan³A

Answer 1

Dear student

$$\begin{gathered} \mathrm{Given} \\ \sec \mathrm{A}=\frac{5}{4}=\frac{\mathrm{Hypotenuse}}{\mathrm{Base}} \\ \mathrm{By}\mathrm{Pythagoras}\mathrm{theorem},\mathrm{we}\mathrm{have} \\ {\left(\mathrm{Hypotenuse}\right)}^2={\left(\mathrm{Perpendicular}\right)}^2+{\left(\mathrm{Base}\right)}^2 \\ \Rightarrow {\left(\mathrm{Perpendicular}\right)}^2=5^2-4^2 \\ \Rightarrow \mathrm{Perpendicular}=3 \\ \mathrm{Now},\mathrm{sinA}=\frac{\mathrm{Perpendicular}}{\mathrm{Hypotenuse}}=\frac{3}{5} \\ \mathrm{cosA}=\frac{\mathrm{Base}}{\mathrm{Hypotenuse}}=\frac{4}{5} \\ \mathrm{TanA}=\frac{\mathrm{Perpendicular}}{\mathrm{Base}}=\frac{3}{4} \\ \mathrm{According}\mathrm{to}\mathrm{ques} \\ \mathrm{LHS}=\frac{3\mathrm{sinA}-4{\sin }^3\mathrm{A}}{4{\cos }^3\mathrm{A}-3\mathrm{cosA}} \\ =\frac{3\left({\displaystyle \frac{3}{5}}\right)-4{\left({\displaystyle \frac{3}{5}}\right)}^3}{4{\left({\displaystyle \frac{4}{5}}\right)}^3-3\left({\displaystyle \frac{4}{5}}\right)} \\ =\frac{{\displaystyle \frac{9}{5}-\frac{108}{125}}}{{\displaystyle \frac{256}{125}-\frac{12}{5}}} \\ =\frac{{\displaystyle \frac{225-108}{125}}}{{\displaystyle \frac{256-300}{125}}} \\ =-\frac{117}{44} \\ \mathrm{RHS}=\frac{3\mathrm{tanA}-{\tan }^3\mathrm{A}}{1-4{\tan }^3\mathrm{A}} \\ =\frac{3\left({\displaystyle \frac{3}{4}}\right)-{\left({\displaystyle \frac{3}{4}}\right)}^3}{1-4{\left({\displaystyle \frac{3}{4}}\right)}^3} \\ =\frac{{\displaystyle \frac{9}{4}-\frac{27}{64}}}{1-{\displaystyle \frac{108}{64}}} \\ =\frac{{\displaystyle \frac{144-27}{64}}}{{\displaystyle \frac{64-108}{64}}} \\ =-\frac{117}{44} \\ \therefore \mathrm{LHS}=\mathrm{RHS} \\ \mathrm{Regards} \end{gathered}$$