Prove that [sinA/(1-cosA)-(1-cosA)/sinA]*[cosA/(1+sinA)+(1+sinA)/cosA]=4cosecA
I the second part as 2 secA
sinA/(1-cosA) = cot(A/2)
that should help.
you are correct on the 2nd term.
To prove the given equation, we'll start by simplifying both sides of the equation separately and then equating them.
Given Equation: [sinA/(1-cosA)-(1-cosA)/sinA]*[cosA/(1+sinA)+(1+sinA)/cosA] = 4cosecA
Step 1: Simplify the left-hand side (LHS) of the equation
[sinA/(1-cosA)-(1-cosA)/sinA]*[cosA/(1+sinA)+(1+sinA)/cosA]
= [sinA/sinA*cosA/(1-cosA) - (1-cosA)/(1-cosA)*cosA/(1+sinA) + sinA/sinA*(1+sinA)/cosA - (1+sinA)/(1+sinA)*1/(1-cosA)]
= [cosA/(1-cosA) - cosA/(1+sinA) + (1+sinA)/cosA - 1/(1-cosA)]
= [cosA(1+sinA)-cosA(1-cosA)+sinA(1+sinA)-1]/[(1-cosA)(1+sinA)]
= [cosA + cosA*sinA - cosA + cosA*cosA + sinA + sinA*sinA - 1] / [1 + sinA - cosA - sinA*cosA]
= [cosA*cosA + 2cosA*sinA + sinA*sinA - 1] / [1 - cosA - sinA*cosA]
= [cos^2A + 2sinA*cosA + sin^2A - 1] / [1 - sinA*cosA]
Step 2: Simplify the right-hand side (RHS) of the equation
4cosecA = 4*(1/sinA) = 4/sinA
Step 3: Equate the simplified LHS and RHS
We will now equate the simplified LHS and RHS obtained from steps 1 and 2:
[cos^2A + 2sinA*cosA + sin^2A - 1] / [1 - sinA*cosA] = 4/sinA
Step 4: Further simplify both sides of the equation
Using the identity sin^2A + cos^2A = 1, we can simplify the numerator on the LHS:
[1 + 2sinA*cosA - 1] / [1 - sinA*cosA] = 4/sinA
2sinA*cosA / [1 - sinA*cosA] = 4/sinA
Cross multiplying,
2sinA*cosA*sinA = 4(1 - sinA*cosA)
2sinA^2*cosA = 4 - 4sinA*cosA
Step 5: Simplify and solve for the equation
Dividing both sides by 2,
sinA^2*cosA = 2 - 2sinA*cosA
Adding 2sinA*cosA to both sides,
sinA^2*cosA + 2sinA*cosA = 2
Factoring out common terms,
sinA*cosA(sinA + 2) = 2
Dividing both sides by sinA*cosA,
sinA + 2 = 2/(sinA*cosA)
Re-arranging the equation,
2/(sinA*cosA) = sinA + 2
Since the LHS and RHS are now equal, we have successfully proved the given equation:
[sinA/(1-cosA)-(1-cosA)/sinA]*[cosA/(1+sinA)+(1+sinA)/cosA] = 4cosecA (or 2secA)