Question

Find the value of $\cot 2A + \tan A = ?$

Answer 1

Hint: In this question we have to use the formula $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$ to do this question. First, we have to convert $\cot 2A = \dfrac{{\cos 2A}}{{\sin 2A}}$ and $\tan A = \dfrac{{\sin A}}{{\cos A}}$ , then we have to apply the above formula to get to the final result.
Formula used: $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$

Complete step-by-step solution:
In the above question,
$ \Rightarrow \cot 2A + \tan A$
Now, use the formula $\cot 2A = \dfrac{{\cos 2A}}{{\sin 2A}}$ and $\tan A = \dfrac{{\sin A}}{{\cos A}}$ in the above equation.
$= \dfrac{{\cos 2A}}{{\sin 2A}} + \dfrac{{\sin A}}{{\cos A}}$
Now taking LCM in the above equation
$=\dfrac{{\cos 2A\cos A + \sin A\sin 2A}}{{\sin 2A\cos A}}$
Now use the identity $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$ in the numerator part
$= \dfrac{{\cos \left( {2A - A} \right)}}{{\sin 2A\cos A}}$
$= \dfrac{{\cos A}}{{\sin 2A\cos A}}$
Now divide numerator and denominator part by $\cos A$
$= \dfrac{1}{{\sin 2A}}$
We know that $\sin \theta = \dfrac{1}{{\cos ec\theta }}$ . So, using this formula in the above equation.
$=\cos ec2A$
Therefore, the answer to the given question is $\cos ec2A$.
Additional information: Trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation. Identity inequalities which are true for every value occurring on both sides of an equation. Geometrically, these identities involve certain functions of one or more angles. There are various distinct identities involving the side length as well as the angle of a triangle. The trigonometric identities hold true only for the right-angle triangle. The six basic trigonometric ratios are sine, cosine, tangent, cosecant, secant, and cotangent. All these trigonometric ratios are defined using the sides of the right triangle, such as an adjacent side, opposite side, and hypotenuse side. All the fundamental trigonometric identities are derived from the six trigonometric ratios.

Note: It is always suggested to convert all the trigonometric functions in terms of sine and cosine functions because they are easiest to solve and there are many identities these functions possess. So there are high chances of solving the question.