Step-by-step explanation:
8) Consider,
[tex] \frac{1}{1 + \sin(x) } [/tex]
Multiplying And Dividing By 1-sinx
[tex] \frac{1}{1 + \sin(x) } \times \frac{1 - \sin(x) }{1 - \sin(x) } [/tex]
[tex] \frac{1 - \sin(x) }{(1 + \sin(x) )(1 - \sin(x)) } [/tex]
We know (a+b)(a-b)=a²-b²
[tex] \frac{1 - \sin(x) }{1 - \sin ^{2} (x) } [/tex]
We know that sin²x+cos²x=1
cos²x=1-sin²x,
uisng this we get,
[tex] \frac{1 - \sin(x) }{ \cos ^{2} (x) } [/tex]
[tex] \frac{1}{ \cos ^{2} (x) } - \frac{ \sin(x) }{cos ^{2} (x)} [/tex]
We know that 1/cosx=secx and sinx/cosx=tanx
Using this we get,
[tex] \sec {}^{2} (x) - \frac{ \tan(x) }{ \cos(x) } [/tex]
Hence Proved
