Respuesta :
[tex]-tan^x + sec^2x = 1[/tex] is proved
Solution:
We have to prove that,
[tex]-tan^x + sec^2x = 1[/tex]
By the trignometric identity,
[tex]sin^2x + cos^2 x = 1[/tex]
Divide both the sides by [tex]cos^2x[/tex] in above identity,
[tex]\frac{sin^2x}{cos^2x} + \frac{cos^2x}{cos^2x} = \frac{1}{cos^2x}[/tex] --- eqn 1
We know that by definition of tan,
[tex]tan x = \frac{sinx}{cosx}[/tex]
Therefore,
[tex]tan^2x = \frac{sin^2x}{cos^2x}[/tex]
Apply the above in eqn 1
[tex]tan^2x + 1 = \frac{1}{cos^2x}[/tex] ---- eqn 2
By definition of cosine,
[tex]cosx = \frac{1}{secx}[/tex]
Therefore,
[tex]cos^2x = \frac{1}{sec^2x}[/tex]
Apply the above in eqn 2
[tex]tan^2x + 1 = sec^2x[/tex]
On rewriting we get,
[tex]sec^2x - tan^2x = 1\\[/tex]
Thus the given identity is proved step by step
