tan x(sin x + cot x cos x) = sec x?

Question

Proving Identities:

tan x(sin x + cot x cos x) = sec x

please answer!

Answer 1

Remember tan(x) = sin(x) / cos(x), etc., so here on the left we have

tan(x) ( sin(x) + cot(x)cos(x) )
[ sin(x) / cos(x) ] ( sin(x) + cot(x)cos(x) )
sec(x) sin(x) ( sin(x) + cot(x)cos(x) )
sec(x) sin(x) ( sin(x) + (cos(x)/sin(x))cos(x) )
sec(x) ( sin^2(x) + cos^2(x))
sec(x) ( 1 )
sec(x)

Answer 2

Convert everything in terms of sin x and cos x:

tan x * (sin x + cot x cos x)

= (sin x / cos x) * (sin x + cos^2 x / sin x)

= (sin^2 x / cos x) + cos x

= (sin^2 x + cos^2 x) / cos x

= 1 / cos x

= sec x

Answer 3

One trick is to bring everything to sin and cos, using
tan = sin/cos
cot = cos/sin
sec = 1/cos

(Sin/Cos)(Sin + CosCos/Sin)

multiplying out:
SinSin/cos + SinCosCos/SinCos

Simplifying (where possible):

SinSin/cos + Cos

Putting everything on same denomicanor (Cos) using
Cos = CosCos/Cos

(1/cos)(Sin^2 + Cos^2)

Use Sin^2 + Cos^2 =1

(1/Cos)(1) = 1/Cos = Sec

Answer 4

Here's what I did for 7]. By parts: let u = ln(2x) and dv = 1/x^2 so that du = 1/(2x) * (2) = 1/x and v = -1/x. Then, int(u dv) = uv - int(v du) = ln(2x) * -1/ x - int(-1/x*1/x) = -ln(2x) / x + int(1/x^2) = -ln(2x) / x - 1/x + C Here's what I did for (3). It may be a little unorthodox... int( (2x-1) / (x + 3)^1/2 ) = int( (2x+6) / (x + 3)^1/2 ) - int( 7 / (x + 3)^1/2 ) (since 6 + (-7) = -1 and breaking into two separate integrals) = 2*int( (x+3) / (x + 3)^1/2 ) - int( 7 / (x + 3)^1/2 ) (factor out 2) = 2*int( (x+3)^1/2 ) - int( 7 / (x + 3)^1/2 ) (simplify (x+3) / (x + 3)^1/2 ) From here, it's just simple u-substitution (let u = x + 3 and integrate. So for example, int( (x+3)^1/2) = 2/3*(x+3)^3/2 ). Thus, we have: = 2*2/3*(x+3)^3/2 - 14(x+3)^1/2 + C = 4/3*(x+3)^3/2 - 14(x+3)^1/2 + C Hope that helps.

Answer 5

first convert cot and cos
cos/sin times cosx/1= cos^2x/sinx add that to sinx
sin^2x+cos^2x/sinx- identity cos^2+sin^2=1
so 1/sinx times tanx
sin/cos times 1/sin = 1/cos= sec