Question

Express \[\cos \theta \] in terms of \[\cot \theta \].

Answer 1

Hint: Here, we have to find the trigonometric ratio in terms of the other. The cosine of an angle is the ratio of the adjacent side to the hypotenuse, where theta is one of the acute angles. If the length of the adjacent gets divided by the length of the opposite side, it becomes the cotangent of an angle in a right triangle.

Formula Used:
We will use the following formulas:
Trigonometric Identity: \[1 + {\tan ^2}\theta = {\sec ^2}\theta \]
Trigonometric Ratio: \[\tan \theta = \dfrac{1}{{\cot \theta }}\]; \[\cos \theta = \dfrac{1}{{\sec \theta }}\];
Law of Surds: \[\sqrt {\dfrac{a}{b}} = \dfrac{{\sqrt a }}{{\sqrt b }}\]; \[\sqrt {{a^2}} = a\]

Complete step-by-step answer:
We will use the trigonometric Identity: \[1 + {\tan ^2}\theta = {\sec ^2}\theta \]
Taking square root on both the sides of above identity, we get
\[ \Rightarrow \sec \theta = \sqrt {1 + {{\tan }^2}\theta } \]
Now, by using the trigonometric ratio \[\tan \theta = \dfrac{1}{{\cot \theta }}\] and substituting it in the above equation, we get
\[ \Rightarrow \sec \theta = \sqrt {1 + \dfrac{1}{{{{\cot }^2}\theta }}} \]
By taking LCM on the right hand side of the above equation, we get
\[ \Rightarrow \sec \theta = \sqrt {1 \times \dfrac{{{{\cot }^2}\theta }}{{{{\cot }^2}\theta }} + \dfrac{1}{{{{\cot }^2}\theta }}} \]
Adding the like terms, we get
\[ \Rightarrow \sec \theta = \sqrt {\dfrac{{{{\cot }^2}\theta + 1}}{{{{\cot }^2}\theta }}} \]
Now, by using the law of Surds \[\sqrt {\dfrac{a}{b}} = \dfrac{{\sqrt a }}{{\sqrt b }}\], we can write
\[ \Rightarrow \sec \theta = \dfrac{{\sqrt {{{\cot }^2}\theta + 1} }}{{\sqrt {{{\cot }^2}\theta } }}\]
Now using the relation \[\cos e{c^2}\theta = 1 + {\cot ^2}\theta \], we get
\[ \Rightarrow \sec \theta = \dfrac{{\sqrt {\cos e{c^2}\theta } }}{{\sqrt {{{\cot }^2}\theta } }}\]
Now, by using the law of Surds \[\sqrt {{a^2}} = a\] , we get
\[ \Rightarrow \sec \theta = \dfrac{{\cos ec\theta }}{{\cot \theta }}\]
Again, by using the trigonometric ratio \[\cos \theta = \dfrac{1}{{\sec \theta }}\], we get
\[ \Rightarrow \dfrac{1}{{\cos \theta }} = \dfrac{{\cos ec\theta }}{{\cot \theta }}\]
\[ \Rightarrow \cos \theta = \dfrac{1}{{\dfrac{{\cos ec\theta }}{{\cot \theta }}}}\]
Rewriting the equation, we get
\[ \Rightarrow \cos \theta = \dfrac{{\cot \theta }}{{\cos ec\theta }}\]
Using the reciprocal relation \[\sin \theta = \dfrac{1}{{\cos ec\theta }}\], we get
\[ \Rightarrow \cos \theta = \cot \theta \cdot \sin \theta \]
Therefore, \[\cos \theta \] can be expressed in terms of \[\cot \theta \] as \[\cos \theta = \cot \theta \cdot \sin \theta \].

Note: We know that the trigonometric ratios of a triangle are also known as trigonometric functions. These are real functions that relate an angle of a right-angled triangle to ratios of a Right-angled triangle. It is possible to convert every trigonometric ratio in terms of the other trigonometric ratio using fundamental identities.
We can also solve by other method such that
\[\cos \theta = \cos \theta \times \dfrac{{\sin \theta }}{{\sin \theta }}\]
By rewriting the equation, we get
\[ \Rightarrow \cos \theta = \dfrac{{\cos \theta }}{{\sin \theta }} \cdot \sin \theta \]
We know that \[\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta \]
\[ \Rightarrow \cos \theta = \cot \theta \cdot \sin \theta \]