Respuesta :
Answer:
The expression [tex]\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}[/tex]
Solution:
From question, given that [tex]\bold{(\sec x+\tan x)^{2}}[/tex]
By using the trigonometric identity [tex](a + b)^{2} = a^{2} + 2ab + b^{2}[/tex] the above equation becomes,
[tex](\sec x+\tan x)^{2} = \sec ^{2} x+2 \sec x \tan x+\tan ^{2} x[/tex]
We know that [tex]\sec x=\frac{1}{\cos x} ; \tan x=\frac{\sin x}{\cos x}[/tex]
[tex](\sec x+\tan x)^{2}=\frac{1}{\cos ^{2} x}+2 \frac{1}{\cos x} \frac{\sin x}{\cos x}+\frac{\sin ^{2} x}{\cos ^{2} x}[/tex]
[tex]=\frac{1}{\cos ^{2} x}+\frac{2 \sin x}{\cos ^{2} x}+\frac{\sin ^{2} x}{\cos ^{2} x}[/tex]
On simplication we get
[tex]=\frac{1+2 \sin x+\sin ^{2} x}{\cos ^{2} x}[/tex]
By using the trigonometric identity [tex]\cos ^{2} x=1-\sin ^{2} x[/tex] ,the above equation becomes
[tex]=\frac{1+2 \sin x+\sin ^{2} x}{1-\sin ^{2} x}[/tex]
By using the trigonometric identity [tex](a+b)^{2}=a^{2}+2ab+b^{2}[/tex]
we get [tex]1+2 \sin x+\sin ^{2} x=(1+\sin x)^{2}[/tex]
[tex]=\frac{(1+\sin x)^{2}}{1-\sin ^{2} x}[/tex]
[tex]=\frac{(1+\sin x)(1+\sin x)}{1-\sin ^{2} x}[/tex]
By using the trigonometric identity [tex]a^{2}-b^{2}=(a+b)(a-b)[/tex] we get [tex]1-\sin ^{2} x=(1+\sin x)(1-\sin x)[/tex]
[tex]=\frac{(1+\sin x)(1+\sin x)}{(1+\sin x)(1-\sin x)}[/tex]
[tex]= \frac{1+\sin x}{1-\sin x}[/tex]
Hence the expression [tex]\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}[/tex]
