Question

State true or false.

If $\cos nx=\sec \theta ,$ then $tan{n}^2\frac{x}{2}={\tan }^2\frac{\theta }{2}$

• A. True
• B. False

Answer 1

The correct option is B False

We know that $\cos 2\theta =\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}$

$\Rightarrow \cos nx=\sec \theta$

$\Rightarrow \cos nx=\frac{1}{\cos \theta }$

$\Rightarrow \frac{1-{\tan }^2\frac{nx}{2}}{1+{\tan }^2\frac{nx}{2}}=\frac{1+{\tan }^2\frac{\theta }{2}}{1-{\tan }^2\frac{\theta }{2}}$

Using componendo-dividendo rule,

$\Rightarrow \frac{1-{\tan }^2\frac{nx}{2}-1-{\tan }^2\frac{nx}{2}}{1-{\tan }^2\frac{nx}{2}+1+{\tan }^2\frac{nx}{2}}=\frac{1+{\tan }^2\frac{\theta }{2}-1+{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}+1-{\tan }^2\frac{\theta }{2}}$

$\Rightarrow \frac{-2{\tan }^2\frac{nx}{2}}{2}=\frac{2{\tan }^2\frac{\theta }{2}}{2}$

$\Rightarrow -{\tan }^2\frac{nx}{2}={\tan }^2\frac{\theta }{2}$