Question

HELP, Prove: (tan(x)+sec(x))^2=2sec^2x+2tansec(x)-1

Answer 1

Expand the left side:

$(\tan x+\sec x)^2=\tan^2x+2\tan x\sec x+\sec^2x$

The Pythagorean identity tells us

$\tan^2x+1=\sec^2x$

so we get

$(\tan x+\sec x)^2=(\sec^2x-1)+2\tan x\sec x+\sec^2x=2\sec^2x+2\tan x\sec x-1$

as needed.

Answer 2

Answer:

(tan(x)+sec(x))²= 2sec²(x)+2tan(x)sec(x)-1.

Step-by-step explanation:

We have given a trigonometric equation:

(tan(x)+sec(x))²=2sec²x+2tansec(x)-1

We have to prove this.

Taking L.H.S  we get,

(tan(x)+sec(x))²= tan²(x)+sec²(x)+2tan(x)sec(x)

Sine we know that :

1+tan²(x) = sec²(x)

Putting in above equation we get,

(tan(x)+sec(x))²= (sec²(x)-1)+sec²(x)+2tan(x)sec(x)

(tan(x)+sec(x))² = 2sec²(x)-1+2tan(x)sec(x)

(tan(x)+sec(x))²= 2sec²(x)+2tan(x)sec(x)-1 = R.H.S

Hence, proved.