Prove $cos(x)+sin(x)tan(x)=sec(x)$ ?

Question

6410 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-22T12:27:04

$cos(x)+sin(x)sin(x)/cos(x)$ (since $tan x =sinx/cosx$ )

$cos(x)+(sin^2x/cosx)$
Taking lcm,
$(cos^2x + sin^2x)/cosx$
$=1/cos (x)$ (since $cos^2x+sin^2x =1$ )
$=secx$
Hence proved

Answer 2 · 2018-01-22T12:27:06

RHS:

$secx=1/cosx$

LHS:

$cosx+sinxtanx$

$=cosx+sinx(sinx/cosx)$

$=cosx+(sin^2x)/cosx$

$=(cosx*cosx)/(1*cosx)+(sin^2x)/cosx$

$=(cos^2x)/(cosx)+(sin^2x)/cosx$

$=(cos^2x+sin^2x)/cosx$

$sin^2x+cos^2x-=1$

$=1/cosx$

$=secx$

$LHS=RHS$

Prove cos(x)+sin(x)tan(x)=sec(x)cos(x)+sin(x)tan(x)=sec(x)?