Prove that: \[\cot A + \tan A = \sec A\cos ecA\]
Answer
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Hint:
The given equation is a trigonometric equation. We will use the formulas of inverse ratios to convert the left-hand side equation into a sum of squares. We will then use the relation between \[\tan A\] and \[\sec A\] to prove the given equation.
Formulas used:
We will use following formulas:
1) \[\cot A = \dfrac{1}{{\tan A}}\]
2) \[\cos ecA = \dfrac{1}{{\sin A}}\]
3) \[\sec A = \dfrac{1}{{\cos A}}\]
4) \[1 + {\tan ^2}A = {\sec ^2}A\]
Complete step by step solution:
We have to prove the given trigonometric equation.
We will first consider the left hand side of the equation.
Now we will use the reciprocal trigonometric ratio.
Substituting \[\cot A = \dfrac{1}{{\tan A}}\] in the left hand side of the given equation, we get
\[\cot A + \tan A = \dfrac{1}{{\tan A}} + \tan A\]
Taking LCM on the right hand side of the above equation, we get
\[ \Rightarrow \cot A + \tan A = \dfrac{{1 + {{\tan }^2}A}}{{\tan A}}\]
We know that \[1 + {\tan ^2}A = {\sec ^2}A\]. Therefore, substituting \[1 + {\tan ^2}A = {\sec ^2}A\] in the above equation, we get
\[\begin{array}{l} \Rightarrow \cot A + \tan A = \dfrac{{{{\sec }^2}A}}{{\tan A}}\\ \Rightarrow \cot A + \tan A = \dfrac{{\sec A \cdot \sec A}}{{\tan A}}\end{array}\]
Now substituting \[\sec A = \dfrac{1}{{\cos A}}\] in the above equation, we get
\[ \Rightarrow \cot A + \tan A = \dfrac{{\sec A}}{{\cos A\tan A}}\]
Substituting \[\tan A = \dfrac{{\sin A}}{{\cos A}}\] in the above equation, we get
\[ \Rightarrow \cot A + \tan A = \dfrac{{\sec A}}{{\cos A \cdot \dfrac{{\sin A}}{{\cos A}}}}\]
Simplifying the above equation, we get
\[ \Rightarrow \cot A + \tan A = \dfrac{{\sec A}}{{\sin A}}\]
Substituting \[\cos ecA = \dfrac{1}{{\sin A}}\] in the above equation, we get
\[ \Rightarrow \cot A + \tan A = \sec A \cdot \cos ecA\]
Thus we proved \[\cot A + \tan A = \sec A\cos ecA\].
Note:
Here, we need to keep in mind the relation between two trigonometric ratios. There are six basic trigonometric ratios and they are sine, cosine, tangent, cosecant, secant and cotangent.
Trigonometry is a branch of mathematics which helps us to study the relationship between the sides and the angles of a triangle. In practical life, cartographers (to make maps) use trigonometry. It is also used by the aviation and naval industries. In fact, trigonometry is even used by Astronomers to find the distance between two stars. Hence, it has an important role to play in everyday life.
The given equation is a trigonometric equation. We will use the formulas of inverse ratios to convert the left-hand side equation into a sum of squares. We will then use the relation between \[\tan A\] and \[\sec A\] to prove the given equation.
Formulas used:
We will use following formulas:
1) \[\cot A = \dfrac{1}{{\tan A}}\]
2) \[\cos ecA = \dfrac{1}{{\sin A}}\]
3) \[\sec A = \dfrac{1}{{\cos A}}\]
4) \[1 + {\tan ^2}A = {\sec ^2}A\]
Complete step by step solution:
We have to prove the given trigonometric equation.
We will first consider the left hand side of the equation.
Now we will use the reciprocal trigonometric ratio.
Substituting \[\cot A = \dfrac{1}{{\tan A}}\] in the left hand side of the given equation, we get
\[\cot A + \tan A = \dfrac{1}{{\tan A}} + \tan A\]
Taking LCM on the right hand side of the above equation, we get
\[ \Rightarrow \cot A + \tan A = \dfrac{{1 + {{\tan }^2}A}}{{\tan A}}\]
We know that \[1 + {\tan ^2}A = {\sec ^2}A\]. Therefore, substituting \[1 + {\tan ^2}A = {\sec ^2}A\] in the above equation, we get
\[\begin{array}{l} \Rightarrow \cot A + \tan A = \dfrac{{{{\sec }^2}A}}{{\tan A}}\\ \Rightarrow \cot A + \tan A = \dfrac{{\sec A \cdot \sec A}}{{\tan A}}\end{array}\]
Now substituting \[\sec A = \dfrac{1}{{\cos A}}\] in the above equation, we get
\[ \Rightarrow \cot A + \tan A = \dfrac{{\sec A}}{{\cos A\tan A}}\]
Substituting \[\tan A = \dfrac{{\sin A}}{{\cos A}}\] in the above equation, we get
\[ \Rightarrow \cot A + \tan A = \dfrac{{\sec A}}{{\cos A \cdot \dfrac{{\sin A}}{{\cos A}}}}\]
Simplifying the above equation, we get
\[ \Rightarrow \cot A + \tan A = \dfrac{{\sec A}}{{\sin A}}\]
Substituting \[\cos ecA = \dfrac{1}{{\sin A}}\] in the above equation, we get
\[ \Rightarrow \cot A + \tan A = \sec A \cdot \cos ecA\]
Thus we proved \[\cot A + \tan A = \sec A\cos ecA\].
Note:
Here, we need to keep in mind the relation between two trigonometric ratios. There are six basic trigonometric ratios and they are sine, cosine, tangent, cosecant, secant and cotangent.
Trigonometry is a branch of mathematics which helps us to study the relationship between the sides and the angles of a triangle. In practical life, cartographers (to make maps) use trigonometry. It is also used by the aviation and naval industries. In fact, trigonometry is even used by Astronomers to find the distance between two stars. Hence, it has an important role to play in everyday life.
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