Question

Sec x -tan x sin x =1/secx

Help me prove it

Answer 1

Answer:

L.H.S. = $\frac{1}{cosx} -\frac{sinx}{cosx}sinx$

= $\frac{1}{cosx} - \frac{sin^2x}{cosx}$

= $\frac{1-sin^x}{cosx} = \frac{cos^2x}{cos x}$

= cos x = 1/secx

Answer 2

Step-by-step explanation:

Hey there!

Given;

sec x - tan x . sin x = 1/(sec x).

~ Taking LHS.

= sec x - tan x . sin x

= 1/(cos x) - (sin x/cos x) . sin x { Using formula sec x = 1/cos x and tan x = (sin x/cos x)}

~ Multiply (sin x) and (sin x).

$= \frac{1 - { (\sin(x) )}^{2} }{ \cos(x) }$

~ Use formula 1-sin²x= cos²x

$= \frac{ {cos}^{2} x}{ \cos(x) }$

~ Cancel cos x.

= cos x

= 1/sec x { cos x = 1/sec x}

→ RHS proved.

Hope it helps.....