Question

Express the trigonometric ratios $\sin A$, $\sec A$ and $\tan A$ in terms of $\cot A$.

Answer 1

Hint:
We can write $\tan A$ as the reciprocal of $\cot A$ . Then we can find $\cos ecA$ using the identity $\cos e{c^2}A = 1 + {\cot ^2}A$ and take its reciprocal to get $\sin A$. Then we can use the identity ${\sec ^2}A = 1 + {\tan ^2}A$ and take its square root. On substituting for $\tan A$, we can obtain the required answer.

Complete step by step solution:
Consider the ratio $\tan A$
We know that $\cot A = \dfrac{1}{{\tan A}}$
 $ \Rightarrow \tan A = \dfrac{1}{{\cot A}}$ …. (2)
Now we can find $\cos ecA$
We know that $\cos e{c^2}A = 1 + {\cot ^2}A$
On taking the square root, we get,
 $\cos ecA = \pm \sqrt {1 + {{\cot }^2}A} $
We assume that A is an acute angle and then all the ratios will be positive. Then we need to take only the positive square root.
 $ \Rightarrow \cos ecA = \sqrt {1 + {{\cot }^2}A} $ … (2)
We know that $\cos ecA = \dfrac{1}{{\sin A}}$
On taking the reciprocal on both sides, we get,
 $ \Rightarrow \sin A = \dfrac{1}{{\cos ecA}}$
On substituting equation (2), we get,
 $ \Rightarrow \sin A = \dfrac{1}{{\sqrt {1 + {{\cot }^2}A} }}$ … (3)
Now we need to find $\sec A$ .
We know that ${\sec ^2}A = 1 + {\tan ^2}A$ .
We can take the square root on both sides.
 $ \Rightarrow \sec A = \pm \sqrt {1 + {{\tan }^2}A} $
We assume that A is an acute angle and then all the ratios will be positive. Then we need to take only the positive square root.
 $ \Rightarrow \sec A = \sqrt {1 + {{\tan }^2}A} $
On substituting equation (1), we get,
 $ \Rightarrow \sec A = \sqrt {1 + {{\left( {\dfrac{1}{{\cot A}}} \right)}^2}} $
On simplification, we get,
 $ \Rightarrow \sec A = \sqrt {1 + \dfrac{1}{{{{\cot }^2}A}}} $
On taking the LCM inside the square root, we get,
 $ \Rightarrow \sec A = \sqrt {\dfrac{{{{\cot }^2}A + 1}}{{{{\cot }^2}A}}} $
On further simplification, we get,
 \[ \Rightarrow \sec A = \dfrac{{\sqrt {{{\cot }^2}A + 1} }}{{\cot A}}\]

Now we have the ratios \[\sec A = \dfrac{{\sqrt {{{\cot }^2}A + 1} }}{{\cot A}}\], $\sin A = \dfrac{1}{{\sqrt {1 + {{\cot }^2}A} }}$, $\tan A = \dfrac{1}{{\cot A}}$ in terms of $\cot A$.

Note:
Alternate solution of finding $\sec A$ is given by,
We know that ${\sin ^2}A + {\cos ^2}A = 1$ .
On rearranging, we get,
 $ \Rightarrow {\cos ^2}A = 1 - {\sin ^2}A$
On taking the square root, we get,
 $ \Rightarrow \cos A = \sqrt {1 - {{\sin }^2}A} $
On substituting the value of $\sin A$ from equation (3), we get,
 $ \Rightarrow \cos A = \sqrt {1 - \dfrac{1}{{\sqrt {1 + {{\cot }^2}A} }}} $
On taking the LCM, we get,
 $ \Rightarrow \cos A = \sqrt {\dfrac{{\sqrt {1 + {{\cot }^2}A} - 1}}{{\sqrt {1 + {{\cot }^2}A} }}} $
On taking the reciprocal, we get,
 $ \Rightarrow \sec A = \sqrt {\dfrac{{\sqrt {1 + {{\cot }^2}A} }}{{\sqrt {1 + {{\cot }^2}A} - 1}}} $
Now we have $\sec A$ in some other form.