# Question #e1c7e

Jul 27, 2016

See the Proof given below in Explanation.

#### Explanation:

Let $\cos \theta = x , \Rightarrow \sec \theta = \frac{1}{x}$. hence, given that,

$\cos \theta + \sec \theta = 2 \Rightarrow x + \frac{1}{x} = 2 , \mathmr{and} , {x}^{2} + 1 = 2 x$

$\therefore {x}^{2} - 2 x + 1 = 0 \Rightarrow {\left(x - 1\right)}^{2} = 0 \Rightarrow x = 1$

$\Rightarrow \cos \theta = 1 , \sec \theta = 1$

Hence, ${\cos}^{n} \theta + {\sec}^{n} \theta = {1}^{n} + {1}^{n} = 2$.

Hence, the Proof.

Here, we have proved that, ${\cos}^{n} \theta + {\sec}^{n} \theta = 2$.

But, ${\cos}^{m} \theta + {\sec}^{n} \theta = 2 , w h e r e , m . n \in \mathbb{R}$ is also true!

On the same lines, the following problems can be solved :

$\left(1\right) : \sin \theta + \csc \theta = 2 \Rightarrow {\sin}^{m} \theta + {\csc}^{n} \theta = 2 , m , n \in \mathbb{R}$

$\left(2\right) : \tan x + \cot x = 2 \rightarrow {\tan}^{2016} x + {\cot}^{2017} x = 2$, etc.

Enjoy Maths!