Question

Express the trigonometric ratios $\sin A,\sec A$ and $\tan A$ in terms of $\cot A$.
A. $
  \sin A = \dfrac{1}{{\sqrt {\cot A + 1} }} \\
  \sec A = \dfrac{{\sqrt {{{\cot }^2}A - 1} }}{{\cot A}} \\
  \tan A = \dfrac{1}{{\cot A}} \\
$

B. $
  \sin A = \dfrac{1}{{\sqrt {{{\cot }^2}A + 1} }} \\
  \sec A = \dfrac{{\sqrt {{{\cot }^2}A + 1} }}{{\cot A}} \\
  \tan A = \dfrac{1}{{\cot A}} \\
$

C. $
  \sin A = \dfrac{1}{{\sqrt {\cot A - 1} }} \\
  \sec A = \dfrac{{\sqrt {{{\cot }^2}A + \sin A} }}{{\cot A}} \\
  \tan A = \dfrac{1}{{\cot A}} \\
$

D. None of these

Answer 1

Hint:We will use simple properties like $1 + co{t^2}A = \cos e{c^2}A$ and $1 + {\tan ^2}A = {\sec ^2}A$ in order to solve the question. Then we will also replace some values after simplifying these properties and turn the ratios of $\sin A,\sec A$ and $\tan A$ into $\cot A$ one by one.

Complete step-by-step answer:
First, in case of $\tan A$, we know that-
$ \Rightarrow \tan A = \dfrac{1}{{\cot A}}$
So, this expresses the ratio of $\sin A$ in terms of $\cot A$ very easily.
Now, we will come to our next term i.e. $\sin A$-
$ \to \sin A = \dfrac{1}{{\cos ecA}}.............(1)$
The trigonometric ratio of $\sin A$ is not in terms of $\cot A$ already. So, we will make it in terms of $\cot A$ like this-
As we know that-
$
   \to 1 + {\cot ^2}A = \cos e{c^2}A \\
    \\
   \to \cos e{c^2}A = 1 + {\cot ^2}A \\
    \\
   \to \cos ecA = \pm \sqrt {1 + {{\cot }^2}A} \\
$
Since A is an acute angle and $\cos ecA$ is positive when A is acute. So,
$ \to \cos ecA = \sqrt {1 + {{\cot }^2}A} $
We will put that value of $\cos ecA$ into equation $(1)$, we will get the ratio of $sinA$ in terms of $\cot A$-
$ \Rightarrow \sin A = \dfrac{1}{{\sqrt {1 + {{\cot }^2}A} }}$
The next term is $\sec A$, we know that-
$
   \to 1 + {\tan ^2}A = {\sec ^2}A \\
    \\
   \to {\sec ^2}A = 1 + {\tan ^2}A \\
    \\
   \to \sec A = \pm \sqrt {\left( {1 + {{\tan }^2}A} \right)} \\
$
Here, A is acute and $\sec A$ is positive when A is acute. So,
$ \to \sec A = \sqrt {\left( {1 + {{\tan }^2}A} \right)} $
$ \to \sec A = \sqrt {\left( {1 + \dfrac{1}{{{{\cot }^2}A}}} \right)} $ (as $\tan A = \dfrac{1}{{\cot A}}$)
$
   \to \sec A = \sqrt {\left( {\dfrac{{{{\cot }^2}A + 1}}{{{{\cot }^2}A}}} \right)} \\
    \\
   \Rightarrow \sec A = \dfrac{{\sqrt {{{\cot }^2}A + 1} }}{{\cot A}} \\
$
Hence, the ratio of $\sec A$ in terms of $\cot A$ is $\sec A = \dfrac{{\sqrt {{{\cot }^2}A + 1} }}{{\cot A}}$.
Thus,
$ \Rightarrow \tan A = \dfrac{1}{{\cot A}}$
$ \Rightarrow \sin A = \dfrac{1}{{\sqrt {1 + {{\cot }^2}A} }}$
$ \Rightarrow \sec A = \dfrac{{\sqrt {{{\cot }^2}A + 1} }}{{\cot A}}$
So, option B is the correct option.

Note: Students should remember trigonometric identities like $1 + co{t^2}A = \cos e{c^2}A$ and $1 + {\tan ^2}A = {\sec ^2}A$ and trigonometric formulas for solving these types of questions.By some simple calculations and using trigonometric identities we can also write $\sin A,\sec A$ and $\tan A$ in terms of $\tan A$.