Tan x /(1 +sec x) + (1+sec x) /tan x
Tan x=sin x / cos x
1+ sec x=1 +1/cos x=(cos x+1)/cos x
Therefore:
tan x /(1 +sec x) =(sin x/cos x)/(cos x+1)/cos x=
=(sin x * cos x) / [cos x* (cos x+1)]=sin x /(Cos x+1)
(1+sec x) /tan x=[(cos x+1)/cos x] / (sin x/cos x)=
=[cos x(cos x+1)]/(sin x *cos x)=(cos x+1)/sin x
tan x /(1 +sec x) + (1+sec x) /tan x=
=sin x /(Cos x+1) + (cos x+1)/sin x=
=(sin²x+cos²x+2 cos x+1) / [sin x(cos x+1)]=
Remember: sin²x+cos²x=1⇒ sin²x=1-cos²x
=(1-cos²x+cos²x+2 cos x+1) / [sin x(cos x+1)]=
=2 cos x+2 / [sin x(cos x+1)]=
=2(cos x+1) / [sin x(cos x+1)]=
=2 /sin x
Answer : tan x /(1 +sec x) + (1+sec x) /tan x= 2/sin x
1-1 is 2. 2 over 2 is the same as 1.
Answer:
30 60 90 triangle
the hypotenuse is twice the length of the shorter leg length.
The answer of this equation is 8cm I did the math
