# Prove sinx/secx+tanx-1 + cosx/cosec x+cotx-1 = 1

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We have to prove that: sin x /(sec x + tan x - 1) + cos x /(cosec x + cot x - 1) = 1

sin x /(sec x + tan x - 1) + cos x /(cosec x + cot x - 1)

=> sin x /(1/cos x + sin x/cos x - 1) + cos x /(1/sin x + cos x/sin x - 1)

=> sin x /(1/cos x + sin x/cos x - cos x/cos x) + cos x /(1/sin x + cos x/sin x - sin x/sin x)

=> sin x * cos x/(1+ sin x - cos x) + cos x*sin x /(1 + cos x - sin x)

=> sin x * cos x[1/(1+ sin x - cos x) + 1/(1 + cos x - sin x)]

=> sin x * cos x[(1 + cos x - sin x) + (1+ sin x - cos x)/(1 + cos x - sin x)(1+ sin x - cos x)]

=> sin x * cos x[(1 + cos x - sin x + 1+ sin x - cos x)/(1^2 - (sin x+ cos x)^2]

=> sin x * cos x[(1 + 1)/(1 - (sin x)^2 - (cos x)^2 + 2*sin x*cos x]

=> sin x * cos x[2/(1 - 1 + 2*sin x*cos x]

=> sin x * cos x[2/2*sin x*cos x]

=> sin x * cos x/sin x*cos x

=> 1

**This proves that sin x /(sec x + tan x - 1) + cos x /(cosec x + cot x - 1) = 1**

If the given expression is sinx/(sec x+tan x-1) + cosx/(cosec x+cot x-1)=1, we'll do the following steps:

sin x/(1/cos x+ sin x/cos x - 1) + cos x/(1/sin x + cos x/sin x - 1) = 1

We've replaced sec x=1/cos x; tan x = sin x/cos x ; cosec x = 1/sin x and cot x = cos x/sin x

sin x/(1/cos x+ sin x/cos x - 1) + cos x/(1/sin x + cos x/sin x - 1) = 1

sin x*cos x/(1+sin x- cos x) + sin x*cos x/(1 + cos x - sin x) = 1

We'll multiply the 1st fraction by (1 + cos x - sin x) and the 2nd fraction by (1+sin x- cos x):

sin x*cos x(1 + cos x - sin x + 1+sin x- cos x)/(1 + cos x - sin x)*(1+sin x- cos x) = 1

We'll eliminate like terms within brackets and we'll multiply the right side by (1 + cos x - sin x)*(1+sin x- cos x):

2 sin x*cos x = (1 + cos x - sin x)*(1+sin x- cos x)

We'll remove the brackets:

2 sin x*cos x = 1+sin x- cos x + cos x + sin x*cos x - (cosx)^2 - sin x - (sin x)^2 + sin x*cos x

But (sin x)^2 + (cosx)^2 = 1

2 sin x*cos x = 1 - 1 + 2sin x*cos x

2 sin x*cos x = 2sin x*cos x

**Since both sides are equal, then the given expression represents an identity.**

thank you :)