Question

If sec theta - tan theta =x show that sec theta 1/2( x+1/x) and tan theta =1/2(1/x-x)

Answer 1

Hey Mate ! Here is your answer.

Answer 2

Answer: Showed

Step-by-step explanation: Given relation is as follows

$\sec \theta-\tan \theta=x.$

We are given to prove the following two results-

$\sec \theta=\dfrac{1}{2}(x+\dfrac{1}{x})$  and  $\tan \theta=\dfrac{1}{2}(\dfrac{1}{x}-x).$  

To prove the above two equalities, we will need the following relation from 'Trigonometry'-

$\sec^{2}\theta=1+\tan^{2}\theta\\\\\Rightarrow \sec^2\theta-\tan^2\theta=1.$

Now,

$\sec \theta=\dfrac{1}{2}(\sec \theta-\tan \theta+\sec \theta+\tan \theta)\\\\\Rightarrow \sec \theta=\dfrac{1}{2}(\sec \theta-\tan \theta+\dfrac{\sec \theta+\tan \theta}{\sec^2\theta-\tan^2\theta})\\\\\Rightarrow \sec \theta=\dfrac{1}{2}(\sec \theta-\tan \theta+\dfrac{1}{\sec \theta-\tan \theta})\\\\\Rightarrow \sec \theta=\dfrac{1}{2}(x+\dfrac{1}{x}),$

and

$\tan \theta=\dfrac{1}{2}(\sec \theta+\tan \theta-\sec \theta+\tan \theta)\\\\\Rightarrow \tan \theta=\dfrac{1}{2}(\dfrac{\sec \theta+\tan \theta}{\sec^2\theta-\tan^2\theta}-\sec \theta+\tan \theta)\\\\\Rightarrow \tan \theta=\dfrac{1}{2}(\dfrac{1}{\sec \theta-\tan \theta}-(\sec \theta-\tan \theta))\\\\\Rightarrow \tan \theta=\dfrac{1}{2}(\dfrac{1}{x}-x).$

Hence proved.