Question

Prove that $\dfrac{{\cos {\text{2A}}}}{{1 - \sin {\text{2A}}}} = \dfrac{{1 + \tan {\text{A}}}}{{1 - \tan {\text{A}}}}$

Answer 1

Hint: We know that the sine and cosine terms can be easily converted into tangent terms, so to prove the given identity convert $\cos {\text{2A}}$ and $\sin {\text{2A}}$ in terms of $\tan {\text{A}}$ by using the respective formulas.

Complete step-by-step answer:
The given identity is $\dfrac{{\cos {\text{2A}}}}{{1 - \sin {\text{2A}}}} = \dfrac{{1 + \tan {\text{A}}}}{{1 - \tan {\text{A}}}}$, we need to prove the given identity. The left hand side part of the identity is in term of $\cos 2{\text{A}}$ and $\sin 2{\text{A}}$, and left hand terms contains only $\tan {\text{A}}$. So, we need to convert $\cos {\text{2A}}$ and $\sin {\text{2A}}$ in terms of $\tan {\text{A}}$.

As we know that e the formula for $\cos {\text{2A}}$ in terms of tangent angle is,
\[\cos {\text{2A}} = \dfrac{{1 - {{\tan }^2}{\text{A}}}}{{1 + {{\tan }^2}{\text{A}}}}\]

Similarly, the formula for $\sin {\text{2A}}$ in terms of tangent angle is,
\[\sin {\text{2A}} = \dfrac{{2\tan {\text{A}}}}{{1 + {{\tan }^2}{\text{A}}}}\]

Now, substitute the value \[\dfrac{{1 - {{\tan }^2}{\text{A}}}}{{1 + {{\tan }^2}{\text{A}}}}\] for $\cos {\text{2A}}$ and \[\dfrac{{2\tan {\text{A}}}}{{1 + {{\tan }^2}{\text{A}}}}\] for $\sin {\text{2A}}$ in the left hand side of the given equation as,
$\dfrac{{\cos {\text{2A}}}}{{1 - \sin {\text{2A}}}} = \dfrac{{\dfrac{{1 - {{\tan }^2}{\text{A}}}}{{1 + {{\tan }^2}{\text{A}}}}}}{{1 - \dfrac{{2\tan {\text{A}}}}{{1 + {{\tan }^2}{\text{A}}}}}}$

Simplify the above equation by taking the LCM of $1 + {\tan ^2}{\text{A}}$ in the denominator and simplify further as,

$
  \dfrac{{\cos {\text{2A}}}}{{1 - \sin {\text{2A}}}} = \dfrac{{\dfrac{{1 - {{\tan }^2}{\text{A}}}}{{1 + {{\tan }^2}{\text{A}}}}}}{{\dfrac{{1 + {{\tan }^2}{\text{A}} - 2\tan {\text{A}}}}{{1 + {{\tan }^2}{\text{A}}}}}} \\
  \dfrac{{\cos {\text{2A}}}}{{1 - \sin {\text{2A}}}} = \dfrac{{1 - {{\tan }^2}{\text{A}}}}{{1 + {{\tan }^2}{\text{A}} - 2\tan {\text{A}}}} \\
 $

As we know the identities,
$
  \left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right) \\
  {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab \\
 $

Use the above identities in the simplified equation as,
$
  \dfrac{{\cos {\text{2A}}}}{{1 - \sin {\text{2A}}}} = \dfrac{{{{\left( 1 \right)}^2} - {{\tan }^2}{\text{A}}}}{{{{\left( 1 \right)}^2} + {{\tan }^2}{\text{A}} - 2\left( 1 \right)\left( {\tan {\text{A}}} \right)}} \\
  \dfrac{{\cos {\text{2A}}}}{{1 - \sin {\text{2A}}}} = \dfrac{{\left( {1 - \tan {\text{A}}} \right)\left( {1 + \tan {\text{A}}} \right)}}{{{{\left( {1 - \tan {\text{A}}} \right)}^2}}} \\
 $

Now, simplify the equation to obtain,

$\dfrac{{\cos {\text{2A}}}}{{1 - \sin {\text{2A}}}} = \dfrac{{\left( {1 + \tan {\text{A}}} \right)}}{{\left( {1 - \tan {\text{A}}} \right)}}$

Hence, it is proved that $\dfrac{{\cos {\text{2A}}}}{{1 - \sin {\text{2A}}}} = \dfrac{{1 + \tan {\text{A}}}}{{1 - \tan {\text{A}}}}$.

Note: While simplifying the equation, arrange the terms of the equation such that we can use the identity of $\left( {{a^2} - {b^2}} \right)$ and the identity of ${\left( {a - b} \right)^2}$ to obtain the desired result and we careful about the sign in the formula of $\cos {\text{2A}}$ and $\sin {\text{2A}}$ in terms of $\tan {\text{A}}$.