Please help me Prove the identity, I keep getting the same answer no matter what

Question

I keep getting the wrong answer and don't kniw how to proceed

Answer 1

$\frac 1 {cosec A-\cot A}=\frac {\sin A} {1-\cos A}=\frac {\sin A (1+\cos A)} {1-\cos^2 A}=\frac {\sin A (1+\cos A)} {sin^2 A}...$

Or, use that $cosec^2 A-\cot^2 A=1 $ ;)

Answer 2

at first the numerator: $$\left(\frac{1}{\cos(A)}-\frac{\sin(A)}{\cos(A)}\right)\left(\frac{1}{\cos(A)}+\frac{\sin(A)}{\cos(A)}\right)=\frac{1}{\cos(A)^2}\left(1-\sin(A))(1+\sin(A)\right))=\frac{\cos(A)^2}{\cos(A)^2}=1$$ and now the denominator: $$\frac{1}{\cos(A)}-\frac{\sin(A)}{\cos(A)}=\frac{1-\sin(A)}{\cos(A)}$$ can you finish?

Answer 3

$$\textrm {L.H.S}=\dfrac {(\sec A - \tan A)(\sec A + \tan A)}{\csc A - \cot A}$$ $$=\dfrac {\sec^2 A - \tan^2 A}{\csc A - \cot A}$$ $$=\dfrac {1}{\csc A - \cot A}$$ $$=\dfrac {\csc^2 A - \cot^2 A}{\csc A - \cot A}$$ $$=\dfrac {(\csc A + \cot A)(\csc A - \cot A)}{(\csc A - \cot A)}$$ $$=\csc A + \cot A=\textrm {R.H.S} \textrm {proved}$$

Answer 4

Eliminate the denominator on the LHS by multiplying by $\csc A - \cot A$. We than get $$(\sec A - \tan A)(\tan A + \sec A) = (\cot A + \csc A)(\csc A - \cot A)$$

Multiply everything on both sides to get $$(\sec^2 A - \tan^2 A)=(\csc^2 A - \cot^2 A)$$

But $(\sec^2 A - \tan^2 A) = 1$ and $(\csc^2 A - \cot^2 A) = 1$ (trigonometric identities!), thus leaving us with $1=1.$ This proves the identity of the LHS and the RHS just as easily (and maybe a little quicker).