`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...
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`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.