Question

Prove that
(Cosec x - sin x)(sec x - cos x)(tan x + cot x) = 1

Answer 1

LHS = (cosec x - sin x)(sec x - cos x)(tan x + cot x) 
    = (1 / sin x - sin x)(1 / cos x - cos x)(tan x + 1 / tan x) 
    = (1 - sin²x)(1 - cos²x)(tan²x + 1) / (sin x * cos x * tan x) 
    = cos²x * sin²x * (tan²x + 1) / (sin x * cos x * tan x), noting sin²x + cos²x = 1 
    = cos²x * sin²x * sec²x / [sin x * cos x * (sin x / cos x)], noting tan²x + 1 = sec²x 
    = sin²x / sin²x, as sec²x = 1 / cos²x 
    = 1 
    = RHS

Answer 2

Answer:

1

Step-by-step explanation:

(cosecx-sinx) (secx-cosx) (tanx+cotx)=1

$(1/sinx-sinx)(1/cosx-cosx)(tanx+1/tanx)$=1

now take L.C.M

So we get,

$(1-sin^2x/sinx)(1-cos^2x/cosx)(1+tan^2x/tanx)=1$

now we know the identities and here we use it

$(cos^2x/sinx)(sin^2x/cosx)(sec^2x/tanx)=1$

$Cosx Sinx(1/cos^2x/tanx)=1\\Cos x Sin x(1/cos^2x*sinx/cosx)=1\\hence prooved$