`sin(x)+cos(x)cot(x) = cosec(x)`

a) To prove this agebraically, you need to draw a right angled triangle.

I will draw the right angled triangle ABC with side ab, and c are the side which are opposite toeir respective angled. For example: side a is opposite of angle A.

Also c is the hypotenuse.

Thus we can write teh LHs of the above expression using algebraic divisions.

`LHS = sin(x)+cos(x)cot(x)`

`= a/c + (b/c)(b/a)`

`= a/c + (b^2)/(ca)`

`= (a^2+b^2)/(ca)`

But according to pythogaras theorem we know,` a^2+b^2 = c^2` .

Therefore,

LHS = (c^2)/(ca)

= c/a

= 1/(a/c) = 1/sin(A)

= cosec(A)

Therefore, from ABC right triangle,

`sin(A)+cos(A)cot(A) = cosec(A)`

Therefore, more generally,

`sin(x)+cos(x)cot(x) = cosec(x)`

b) This can be proved algebraically with the exception of angle is zero where a is zero and when angle is 90 at which a triangle cannot be constructed.