Question

Prove that: \[\sqrt{\dfrac{1-\cos A}{1+\cos A}}=\text{cosec}A-\cot A\]

Answer 1

Hint: We will begin with the left hand side of the equation (1) and then we will solve the right side of the equation (1) and we will get the same answer from both the sides. We will use the formulas \[1-\cos A=2{{\sin }^{2}}\dfrac{A}{2}\] and \[1+\cos A=2{{\cos }^{2}}\dfrac{A}{2}\] to prove the expression.
Complete step-by-step answer:

It is mentioned in the question that \[\sqrt{\dfrac{1-\cos A}{1+\cos A}}=\text{cosec}A-\cot A........(1)\]

Now solving the right hand side of the equation (1) we get,

\[\Rightarrow \sqrt{\dfrac{1-\cos A}{1+\cos A}}.....(2)\]

We know that \[\cos A={{\cos }^{2}}\dfrac{A}{2}-{{\sin }^{2}}\dfrac{A}{2}.......(3)\]
So converting the cos term to sin in the right hand side of the equation (3) we get,

\[\begin{align}

  & \cos A=1-{{\sin }^{2}}\dfrac{A}{2}-{{\sin }^{2}}\dfrac{A}{2} \\

 & \cos A=1-2{{\sin }^{2}}\dfrac{A}{2}......(4) \\

\end{align}\]

Rearranging equation (4) we get,

\[1-\cos A=2{{\sin }^{2}}\dfrac{A}{2}......(5)\]

Similarly converting the sin term to cos in the right hand side of the equation (3) we get,

\[\begin{align}

  & \cos A={{\cos }^{2}}\dfrac{A}{2}-1+{{\cos }^{2}}\dfrac{A}{2} \\

 & \cos A=2{{\cos }^{2}}\dfrac{A}{2}-1........(6) \\

\end{align}\]

Rearranging equation (6) we get,

\[1+\cos A=2{{\cos }^{2}}\dfrac{A}{2}........(7)\]

Now substituting the values of 1-cosA from equation (5) and 1+cosA from equation (7) in equation (2) w get,

\[\Rightarrow \sqrt{\dfrac{2{{\sin }^{2}}\dfrac{A}{2}}{2{{\cos }^{2}}\dfrac{A}{2}}}.......(8)\]

Cancelling similar terms in equation (5) we get,

\[\Rightarrow \sqrt{{{\tan }^{2}}\dfrac{A}{2}}=\tan \dfrac{A}{2}.......(9)\]

Now solving the right hand side of the equation (1) we get,

\[\Rightarrow \text{cosec}A-\cot A.........(10)\]

Simplifying both the terms in equation (10) we get,

\[\Rightarrow \dfrac{1}{\sin A}-\dfrac{\cos A}{\sin A}.........(11)\]

Now taking the LCM in equation (11) and solving we get,

\[\Rightarrow \dfrac{1-\cos A}{\sin A}.........(12)\]

We know that \[\sin A=2\sin \dfrac{A}{2}\cos \dfrac{A}{2}\] so substituting it in equation (12)

and also substituting from equation (5) in equation (12) and solving we get,

\[\Rightarrow \dfrac{2{{\sin }^{2}}\dfrac{A}{2}}{2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}=\dfrac{\sin \dfrac{A}{2}}{\cos \dfrac{A}{2}}=\tan \dfrac{A}{2}......(13)\]

Hence from equation (9) and equation (13) we can see that the answer is the same on solving both sides. Hence we have proved equation (1).

Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We may get confused at first seeing square root but we have to remember the formulas \[1-\cos A=2{{\sin }^{2}}\dfrac{A}{2}\] and \[1+\cos A=2{{\cos }^{2}}\dfrac{A}{2}\] then we can solve this. In equation (10) we need to transform the whole equation in terms of sin and cos.