Question

Prove that cotA + sinA/ cotA - sinA = 1-cosA+secA/1+ cosA-secA

Answer 1

Lets take Left Hand Side 
cosecA (secA-1) -cotA (1-cosA) 

Now let's simplify i.e. make cosecA as 1/sinA , secA as 1/cosA and cotA as cosA/sinA 
we get- 

(1/sinA)[(1/cosA)-1] - (cosA/sinA)(1-cosA) 

Now. let's do cross multiplication of [(1/cosA)-1] 
we get- 

(1/sinA)[(1-cosA)/cosA] - (cosA/sinA)(1-cosA) 

Now, let's multiply brackets right to the - sign as well as left to the - sign 
we get- 

[(1-cosA)/(sinA.cosA)] - [cosA(1-cosA)/sinA] 

Now if you see the part (1-cosA)/sinA is common part so we will take it out and we get- 

(1-cosA)/sinA [(1/cosA) -cosA] 

Again perform cross multiplication inside [(1/cosA) -cosA] 
we get- 

[(1-cosA)/sinA] [(1-cos^2A)/cosA] 

Now, 1-cos^2A=sin^2A 
So, 

[(1-cosA)/sinA] [sin^2A/cosA] 

Now, let's multiply both the square brackets 
we get- 

[(1-cosA).sin^2A]/[sinA.cosA] 

We will cancel one sinA from numerator (sin^2A) and denominator (sinA), we get 

(1-cosA).sinA/cosA 

Now do the multiplication- 

sinA/cosA - cosA*sinA/cosA 

Now, sinA/cosA=tanA and cancel one cosA from numerator and denominator and we get- 

tanA - sinA 

which is our required Right Hand Side.. :-)