[tex]\frac{1+\tan x}{\sin x+\cos x}=\sec x[/tex] is proved
Given that,
[tex]\frac{1+\tan x}{\sin x+\cos x}=\sec x[/tex] ------- (1)
First we will simplify the LHS and then compare it with RHS
[tex]\text { L. H.S }=\frac{1+\tan x}{\sin x+\cos x}[/tex] ------ (2)
[tex]\text {We know that } \tan x=\frac{\sin x}{\cos x}[/tex]
Substitute this in eqn (2)
[tex]=\frac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}[/tex]
On simplification we get,
[tex]=\frac{\frac{\sin x+\cos x}{\cos x}}{\sin x+\cos x}[/tex]
[tex]=\frac{\sin x+\cos x}{\cos x} \times \frac{1}{\sin x+\cos x}[/tex]
Cancelling the common terms (sinx + cosx)
[tex]=\frac{1}{c o s x}[/tex]
We know secant is inverse of cosine
[tex]=\sec x=R . H . S[/tex]
Thus L.H.S = R.H.S
[tex]\frac{1+\tan x}{\sin x+\cos x}=\sec x[/tex]
Hence proved
