Question

tan^2x+sec^2x=1 for all values of x.
true or false?

Answer 1

This is not true.

$\tan^2x+\sec^2x=\dfrac{\sin^2x}{\cos^2x}+\dfrac1{\cos^2x}=\dfrac{1-\cos^2x+1}{\cos^2x}=2\sec^2x-1$

$2\sec^2x-1=1\implies\sec^2x=1\implies\sec x=\pm1\implies x=n\pi$

where is $n$ is any integer. So suppose we pick some value of $x$ other than these, say $x=\dfrac\pi4$. Then

$\tan^2\dfrac\pi4+\sec^2\dfrac\pi4=1+2=3\neq1$

Answer 2

The trigonometrical identity says : 1+tan^2x = sec^2 x that means

sec^2 x - tan^2 x = 1

However in our question, its tan^2 x + sec^2 x or say sec^2 x + tan^2 x = 1 which is completely against the identity.

Hense answer is False.

Additional information: There is, with a little difference, as equation which is true ie tan^2 x + sec x = 1 for some values of x.