**sec(-x)cotx**put secx=1/cosx

and cotx=cosx/sinx

=> -(1/cosx)*(cosx/sinx)

=> -(1/sinx)

put 1/sinx=cosecx

=> -cosecx

So

**Sec(-x)cot x**=

**-cosec x**

Cheers

and cotx=cosx/sinx

=> -(1/cosx)*(cosx/sinx)

=> -(1/sinx)

put 1/sinx=cosecx

=> -cosecx

So

Cheers

Solve 2*sec(x) - 1 = 0?

It is a question of trigonometric functions.

Sec(-x)cot X

since Secx= 1/because x

So,

Sec(-x)cot X= 1/ because(-x) Cot x

as because(-x)= Cosx, So

Sec(-x)cot X=1/Cosx Cotx

Put Cotx= because x/ Sinx, we get

Sec(-x)cot X= 1/Cosx * because x/ Sin x

Sec(-x)cot X=Cosx/ Cosx * 1/Sin x

Sec(-x)cot X= 1* 1/Sinx

Sec(-x)cot X= 1/ Sinx

we know that 1/ Sin x= Cosec x

So,

Sec(-x)cot X= Cosec x

Sec(-x)cot X

since Secx= 1/because x

So,

Sec(-x)cot X= 1/ because(-x) Cot x

as because(-x)= Cosx, So

Sec(-x)cot X=1/Cosx Cotx

Put Cotx= because x/ Sinx, we get

Sec(-x)cot X= 1/Cosx * because x/ Sin x

Sec(-x)cot X=Cosx/ Cosx * 1/Sin x

Sec(-x)cot X= 1* 1/Sinx

Sec(-x)cot X= 1/ Sinx

we know that 1/ Sin x= Cosec x

So,

Sec(-x)cot X= Cosec x

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