Question

Can anybody help me verify this identity?

cos x/(1 + sin x) + (1+sin x/cos x) = 2 sec x

Answer 1

@SithsAndGiggles can you help me with this?

Answer 2

\[\frac{\cos x}{1+\sin x}+\frac{1+\sin x}{\cos x}=2\sec x~~?\]

Answer 3

Yes, that is the identity

Answer 4

I got to cos\[\cos ^{2}x + \sin ^{2} x + 2/1 + \sin x \cos x\]

Answer 5

Is the equation as I wrote it? I was asking for clarification.

Answer 6

Yes, it is

Answer 7

\[\begin{align*}\frac{\cos x}{1+\sin x}+\frac{1+\sin x}{\cos x}&=\frac{\cos x}{1+\sin x}\cdot\frac{\cos x}{\cos x}+\frac{1+\sin x}{\cos x}\cdot\frac{1+\sin x}{1+\sin x }\\\\
&=\frac{\cos^2 x+(1+\sin x)^2}{\cos x(1+\sin x)}\\\\
&=\frac{\cos^2 x+1+2\sin x+\sin^2x}{\cos x(1+\sin x)}\\\\
&=\frac{2+2\sin x}{\cos x(1+\sin x)}\\\\
&=\frac{2(1+\sin x)}{\cos x(1+\sin x)}\\\\
&=\frac{2}{\cos x}\\\\
&=2\sec x
\end{align*}\]

Answer 8

I was about to type that :) @SithsAndGiggles

Answer 9

@SithsAndGiggles how did you get to the 4th step? (The one after expanding the equation)

Answer 10

\[(a+b)^2)=a^2+2ab+b^2\]

Answer 11

\[sin^2(x)+cos^2(x)=1\]

Answer 12

@bakur Thank you!

Answer 13

\[L.H.S=\frac{ \cos x }{ 1+\sin x }\times \frac{ 1-\sin x }{ 1-\sin x }+\frac{ 1+\sin x }{ \cos x }\]
\[=\frac{ \cos x \left( 1-\sin x \right) }{ \left(1 +\sin x \right)\left( 1-\sin x \right) }+\frac{ 1+\sin x }{ \cos x }\]
\[=\frac{ \cos x \left( 1-\sin x \right) }{ 1-\sin ^2x }+\frac{ 1+\sin x }{ \cos x }\]
\[=\frac{ \cos x \left( 1-\sin x \right) }{ \cos ^2x }+\frac{ 1+\sin x }{ \cos x }\]
\[=\frac{ 1-\sin x }{ \cos x }+\frac{ 1+\sin x }{ \cos x }\]
\[=\frac{ 1-\sin x+1+\sin x }{ \cos x }\]
\[=\frac{ 2 }{ \cos x }\]
=2 sec x
=R.H.S