Prove
SinA/1-CosA = CosecA + CotA
LHS:
SinA/(1-cosA) X 1+cosA/1+cosA ( applying the sure method , i.e rationalising)
SinA(1+cosA)/1XCos^2A
SinA(1+cosA)/sin^2A
SinA+sinACosA/Sin^2A
SinA/Sin^2A + SinA+CosA/Sin^A
CosecA+CotA =RHS (proved)
Reminder: ^ symbol means to the power
pl help
In my previous response , where I said the sure method ... It will be surd method , it got autocorrected 🤣
And one more thing , in the second line it will be:
SinA(1+cosA)/1-cos^2A
Sorry for the mistakes guys 😢
To prove that SinA/1-CosA = CosecA + CotA, we can start with the left-hand side (LHS) of the equation and simplify it until we arrive at the right-hand side (RHS) of the equation.
1. Start with LHS: SinA/1-CosA
2. Rewrite SinA as 1/CosecA and CosA as 1/SecA:
(1/CosecA) / [1 - (1/SecA)]
3. Simplify the denominator by multiplying it by SecA/SecA:
(1/CosecA) / [(SecA - 1)/SecA]
4. Invert the denominator to change the division to multiplication:
(1/CosecA) * (SecA/ (SecA - 1))
5. Use the identity CosecA = 1/SinA and simplify:
(1/(1/SinA)) * (SecA/ (SecA - 1))
6. Simplify by multiplying the reciprocals:
SinA * (SecA/ (SecA - 1))
7. Rewrite SecA as 1/CosA:
SinA * (1/CosA)/ (1/CosA - 1)
8. Simplify further by inverting the denominator to change the division to multiplication:
SinA * (1/CosA) * (CosA/(CosA - 1))
9. Simplify by canceling out the common terms:
SinA * 1 * 1/(CosA - 1)
10. Finally, simplify to the RHS:
SinA / (CosA - 1) = CosecA + CotA
Hence, we have proved that SinA/1-CosA = CosecA + CotA by simplifying the left-hand side and arriving at the right-hand side of the equation.
Assuming all are using angle A,
sin/(1-cos)
= sin(1+cos)/(1-cos)(1+cos)
= sin(1+cos)/(1-cos^2)
= sin(1+cos)/sin^2
= (1+cos)/sin
= csc+cot