Question

How do you find a double angle formula for $\sec 2x$ in terms of only $\mathrm{cosec}x$ and $\sec x$?

Answer 1

Determine the formula using trigonometric identities:

The double angle formula for $\sec 2x$ can be determined using trigonometric identities as follows,

$\sec 2x=\frac{1}{\cos 2x}$ $\mathrm{[}\mathrm{\because }\mathrm{}\mathrm{secx}\mathrm{=}\frac{\mathrm{1}}{\mathrm{cosx}}\mathrm{]}$

$\Rightarrow \sec 2x=\frac{1}{\cos \left(x+x\right)}$

$\Rightarrow \sec 2x=\frac{1}{\cos x·\cos x-\sin x·\sin x}$ $\mathrm{[}\mathrm{\because }\mathrm{}\cos \left(A+B\right)\mathrm{}\mathrm{=}\mathrm{}\mathrm{cosA}\mathrm{·}\mathrm{cosB}\mathrm{}\mathrm{-}\mathrm{}\mathrm{sinA}\mathrm{·}\mathrm{sinB}\mathrm{]}$

$\Rightarrow \sec 2x=\frac{1}{{\displaystyle \frac{1}{\sec x·\sec x}}-{\displaystyle \frac{1}{\mathrm{cosec}x·c\mathrm{osec}x}}}$ $\mathrm{[}\mathrm{\because }\mathrm{}\mathrm{cosx}\mathrm{=}\frac{\mathrm{1}}{\mathrm{secx}}\mathrm{,}\mathrm{}\mathrm{sinx}\mathrm{=}\frac{\mathrm{1}}{\mathrm{cosecx}}\mathrm{]}$

$\Rightarrow \sec 2x=\frac{1}{{\displaystyle \frac{1}{{\sec }^{2}x}}-{\displaystyle \frac{1}{{\mathrm{cosec}}^{2}x}}}$

$\Rightarrow \sec 2x=\frac{1}{{\displaystyle \frac{{\mathrm{cosec}}^{2}x-{\sec }^{2}x}{{\sec }^{2}x·{\mathrm{cosec}}^{2}x}}}$

$\Rightarrow \sec 2x=\frac{{\sec }^{2}x·{\mathrm{cosec}}^{2}x}{{\displaystyle {\mathrm{cosec}}^{2}x-{\sec }^{2}x}}$ $\mathrm{[}\mathrm{\because }\mathrm{}\frac{\mathrm{1}}{{\displaystyle \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$b$}\right.}}\mathrm{=}\raisebox{1ex}{$\mathrm{b}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{a}$}\right.\mathrm{]}$

Hence, the double angle formula for $\sec 2x=\frac{{\sec }^{2}x·{\mathrm{cosec}}^{2}x}{{\displaystyle {\mathrm{cosec}}^{2}x-{\sec }^{2}x}}$.