Question

Prove the following:
(iv) $se{c}^{6}x-{\tan }^{6}x=1+3se{c}^{2}x{\tan }^{2}x$

Answer 1

$$\begin{gathered} \mathrm{LHS}={\sec }^{6}x-{\tan }^{6}x \\ ={\left(se{c}^{2}x\right)}^3-{\left({\tan }^{2}x\right)}^3 \\ =\left(se{c}^{2}x-{\tan }^{2}x\right)\left(se{c}^{4}x+se{c}^{2}x{\tan }^{2}x+{\tan }^{4}x\right)\left[\mathrm{Using}{a}^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\right] \\ =1\times \left(se{c}^{4}x+se{c}^{2}x{\tan }^{2}x+{\tan }^{4}x\right)\left(\mathrm{Since},se{c}^{2}x-{\tan }^{2}x=1\right) \\ ={\left(se{c}^{2}x\right)}^2+{\left({\tan }^{2}x\right)}^2+se{c}^{2}x{\tan }^{2}x \\ ={\left(se{c}^{2}x-{\tan }^{2}x\right)}^2+2se{c}^{2}x{\tan }^{2}x+se{c}^{2}x{\tan }^{2}x \\ =1+3se{c}^{2}x{\tan }^{2}x \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$