Now, x^2 2x^2 x+1 =x^2 2x^2 x+(^2 x tan^2 x) [ ^2 x tan^2 x=1 ] =x^2 2x^2 x+^2 x tan^2 x =(x x)^2 tan^2 x =(x sec x tan x)(x x+ta ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Basic Trigonometric Identities

Simplify: x2...

Question

Simplify:
$x^{2} - 2 x {sec}^{2} x + 1$

Open in App

Solution

Now,
$x^{2} - 2 x {sec}^{2} x + 1$
$= x^{2} - 2 x {sec}^{2} x + ({sec}^{2} x - {tan}^{2} x)$ [ ${sec}^{2} x - {tan}^{2} x = 1$ ]
$= x^{2} - 2 x {sec}^{2} x + {sec}^{2} x - {tan}^{2} x$
$= (x - sec x)^{2} - {tan}^{2} x$
$= (x - s e c x - tan x) (x - sec x + tan x)$ .

Suggest Corrections

Similar questions

2 x^{2} (x + 2) - 3 x (x^{2} - 3) - 5 x (x + 5)

2 x (2 x - 3) + 5

x = 2

∣ ∣ ∣ ∣ \begin{matrix} x^{2} + 2 x & 2 x + 1 & 1 2 x + 1 & x + 2 & 1 3 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ = {(x - 1)}^{3} .

(x^{2} - 2 x) (2 x - 2) - 9 \frac{2 x - 2}{x^{2} - 2 x} ⩽ 0

{tan}^{- 1} [\frac{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}] =

Join BYJU'S Learning Program