Prove that tanx/(1+secx)+(1+secx)/tanx=2secx?

Feb 23, 2016

It is not an identity. It cannot be verified.

Explanation:

The left side is not always equal to the right. For example at $x = \frac{\pi}{6}$ the left side is $4$ and the right side is $\frac{4 \sqrt{3}}{3}$

Feb 23, 2016

It should be $\tan \frac{x}{1 + \sec x} + \frac{1 + \sec x}{\tan} x = 2 \csc x$. See proof below.

Explanation:

$\tan \frac{x}{1 + \sec x} + \frac{1 + \sec x}{\tan} x \ne 2 \sec x$. It should rather be $\tan \frac{x}{1 + \sec x} + \frac{1 + \sec x}{\tan} x = 2 \csc x$

Perhaps what you mean is to prove that $\tan \frac{x}{1 + \sec x} + \frac{1 + \sec x}{\tan} x = 2 \csc x$. To solve this let us start from LHS and prove using other identities that this is equal to RHS.

$\tan \frac{x}{1 + \sec x} + \frac{1 + \sec x}{\tan} x$ (let us first add them like fractions)

= (tan^2x+(1+secx)^2)/(tanx(1+secx) (expanding this)

= (tan^2x+1+sec^2x+2secx)/(tanx(1+secx)(using ${\sec}^{2} x = 1 + {\tan}^{2} x$)

= (sec^2x+sec^2x+2secx)/(tanx(1+secx) or

=(2sec^2x+2secx)/(tanx(1+secx) or

= (2secx(secx+1))/(tanx(1+secx)

Now, using $\sec x = \frac{1}{\cos} x$ and $\tan x = \sin \frac{x}{\cos} x$

= $2 \sec \frac{x}{\tan} x$ =$\frac{\frac{2}{\cos} x}{\sin \frac{x}{\cos} x}$ (simplifying & using $\frac{1}{\sin} x = \csc x$)

= $\left(\frac{2}{\sin} x\right)$ = $2 \csc x$