Question

Prove that:
\[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta \]

Answer 1

Hint: By using the trigonometric identities we can convert the cotangent terms into the tangent terms and then simplify it using the corresponding identities.

Complete step-by-step answer:
\[\begin{align}
  & \cot \theta =\dfrac{1}{\tan \theta } \\
 & \sec \theta =\dfrac{1}{\cos \theta } \\
 & \operatorname{cosec}\theta =\dfrac{1}{\sin \theta } \\
\end{align}\]

\[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\]

Relations between different sides and angles of a right angled triangle are called trigonometric ratios.

\[\begin{align}
  & \sin \theta =\dfrac{\text{Opposite}}{\text{Hypotenuse}} \\
 & \cos \theta =\dfrac{\text{adjacent}}{\text{Hypotenuse}} \\
 & \tan \theta =\dfrac{\text{Opposite}}{\text{adjacent}} \\
\end{align}\]

TRIGONOMETRIC IDENTITIES:

An equation involving trigonometric functions which is true for all those angles for which the functions are defined is called trigonometric identity.

\[\begin{align}
  & \cot \theta =\dfrac{1}{\tan \theta } \\
 & \sec \theta =\dfrac{1}{\cos \theta } \\
 & \operatorname{cosec}\theta =\dfrac{1}{\sin \theta } \\
\end{align}\]

\[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\]

Now, from the given equation we get,

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }\]

\[\Rightarrow \dfrac{\tan \theta }{1-\dfrac{1}{\tan \theta }}+\dfrac{\dfrac{1}{\tan \theta

 }}{1-\tan \theta }\text{ }\left[ \because \cot \theta =\dfrac{1}{\tan \theta } \right]\]

\[\Rightarrow \dfrac{{{\tan }^{2}}\theta }{\tan \theta -1}+\dfrac{1}{\tan \theta \left( 1-\tan

\theta \right)}\]

Now, take the common term out and rewrite the equation.

\[\begin{align}
  & \Rightarrow \dfrac{1}{1-\tan \theta }\left( \dfrac{1}{\tan \theta }-{{\tan }^{2}}\theta \right) \\
 & \Rightarrow \dfrac{1}{1-\tan \theta }\left( \dfrac{1-{{\tan }^{3}}\theta }{\tan \theta } \right) \\
\end{align}\]

\[\Rightarrow \dfrac{\left( 1-\tan \theta \right)\left( 1+{{\tan }^{2}}\theta +\tan \theta

\right)}{\left( 1-\tan \theta \right)\tan \theta }\text{ }\left[ \because

{{x}^{3}}-{{a}^{3}}=\left( x-a \right)\left( {{x}^{2}}+{{a}^{2}}+ax \right) \right]\]

\[\Rightarrow \dfrac{{{\sec }^{2}}\theta +\tan \theta }{\tan \theta }\text{ }\left[ \because

 {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 \right]\]

\[\begin{align}
  & \Rightarrow \dfrac{{{\sec }^{2}}\theta }{\tan \theta }+1\text{ }\left[ \because \dfrac{\sec \theta }{\tan \theta }=\operatorname{cosec}\theta \right] \\
 & \Rightarrow 1+\sec \theta \operatorname{cosec}\theta \\
\end{align}\]

Note: Instead of converting the cotangent terms into tangent terms we can also convert them into sine and cosine terms and then simplify the equation so formed using the corresponding trigonometric identities. Both the methods give the same result but this one will be more lengthy.

\[\dfrac{\sin \theta }{\cos \theta }=\tan \theta \]
It is important to note that the following cubic equation formed in tangent terms can be simplified into the product of the linear and quadratic terms which thereby cancels out the common terms. Then on the further simplification of the equation formed using the trigonometric identities gives the result.