Question

Solve the given expression $\tan \left( {65 - \theta } \right) - \cot \left( {25 + \theta } \right)$

Answer 1

Hint –There can be two ways to solve this problem, in the first one we use the concept that $\tan \left( {90 - \theta } \right) = \cot \theta $, so write $\cot \left( {25 + \theta } \right)$ as $\tan \left( {90 - \left( {25 + \theta } \right)} \right)$, to get the answer, and the second method is explained in the later end.
Complete Step-by-Step solution:
Given trigonometric equation is
$\tan \left( {65 - \theta } \right) - \cot \left( {25 + \theta } \right)$
As we know that $\tan \left( {90 - A} \right) = \cot A$ so use this property in above equation we have,
Here $A = \left( {25 + \theta } \right)$.
$ \Rightarrow \tan \left( {65 - \theta } \right) - \cot \left( {25 + \theta } \right) = \tan \left( {65 - \theta } \right) - \tan \left( {90 - \left( {25 + \theta } \right)} \right)$
Now simplify the above equation we have
$ \Rightarrow \tan \left( {65 - \theta } \right) - \cot \left( {25 + \theta } \right) = \tan \left( {65 - \theta } \right) - \tan \left( {90 - 25 - \theta } \right)$
$ \Rightarrow \tan \left( {65 - \theta } \right) - \cot \left( {25 + \theta } \right) = \tan \left( {65 - \theta } \right) - \tan \left( {65 - \theta } \right)$
Now as we both terms are equal and subtracting each other so it will cancel out therefore we have,
$ \Rightarrow \tan \left( {65 - \theta } \right) - \cot \left( {25 + \theta } \right) = 0$
So the required answer of the given trigonometric equation is zero.
So this is the required answer.

Note –  The second method involves the use of basic trigonometric identities that helps conversion of one trigonometric ratios directly into other trigonometric ratio like $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta {\text{ }}}}{\text{ and cot}}\theta {\text{ = }}\dfrac{{\cos \theta }}{{\sin \theta }}$. Further use of basic identities of $\cos (A + B) = \cos A\cos B - \sin A\sin B$ will help getting the answer.