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Hint: Here, we will use some basic formulas of trigonometry which are $\sin A = \dfrac{1}{{{\text{cosec}}A}}$,$\tan A = \dfrac{1}{{\cot A}}$ and some trigonometric identities which are ${\left( {{\text{cosec}}A} \right)^2} = 1 + {\left( {\cot A} \right)^2}$, ${\left( {\sec A} \right)^2} = 1 + {\left( {\tan A} \right)^2}$.

Complete step-by-step answer:

As we know that the sine trigonometric function is the reciprocal of cosine trigonometric function.

i.e., $\sin A = \dfrac{1}{{{\text{cosec}}A}}{\text{ }} \to {\text{(1)}}$

Using the formula ${\left( {{\text{cosec}}A} \right)^2} = 1 + {\left( {\cot A} \right)^2}$, we can

$ \Rightarrow {\text{cosec}}A = \sqrt {1 + {{\left( {\cot A} \right)}^2}} $

Using the above equation, equation (1) becomes

$ \Rightarrow \sin A = \dfrac{1}{{\sqrt {1 + {{\left( {\cot A} \right)}^2}} }}$

This above equation represents the sine trigonometric function in terms of cotangent trigonometric function.

Since, tangent trigonometric function is the reciprocal of cotangent trigonometric function.

Also we know that ${\left( {\sec A} \right)^2} = 1 + {\left( {\tan A} \right)^2}{\text{ }} \to {\text{(2)}}$ and $\tan A = \dfrac{1}{{\cot A}}{\text{ }} \to {\text{(3)}}$

Substituting the value of tangent trigonometric function from equation (3) in equation (2), we get

\[

{\left( {\sec A} \right)^2} = 1 + {\left( {\dfrac{1}{{\cot A}}{\text{ }}} \right)^2} = 1 + \dfrac{1}{{{{\left( {\cot A} \right)}^2}}} \\

\Rightarrow \sec A = \sqrt {1 + \dfrac{1}{{{{\left( {\cot A} \right)}^2}}}} \\

\]

This above equation represents the secant trigonometric function in terms of cotangent trigonometric function.

According to equation (3), we can write

$\tan A = \dfrac{1}{{\cot A}}$

This above equation represents the tangent trigonometric function in terms of cotangent trigonometric function.

Hence, $\sin A = \dfrac{1}{{\sqrt {1 + {{\left( {\cot A} \right)}^2}} }}$, \[\sec A = \sqrt {1 + \dfrac{1}{{{{\left( {\cot A} \right)}^2}}}} \] and $\tan A = \dfrac{1}{{\cot A}}$

Note: In this particular problem, in order to represent $\sin A$ in terms of $\cot A$ we will firstly convert $\sin A$ in terms of ${\text{cosec}}A$ and then finally we will convert ${\text{cosec}}A$ in terms of $\cot A$. Also, in order to convert $\sec A$ in terms of $\cot A$ we will firstly convert $\sec A$ in terms of $\tan A$ and then finally we will convert $\tan A$ in terms of $\cot A$. Also, for representing $\tan A$ in terms of $\cot A$ we will simply use the formula $\tan A = \dfrac{1}{{\cot A}}$.

Complete step-by-step answer:

As we know that the sine trigonometric function is the reciprocal of cosine trigonometric function.

i.e., $\sin A = \dfrac{1}{{{\text{cosec}}A}}{\text{ }} \to {\text{(1)}}$

Using the formula ${\left( {{\text{cosec}}A} \right)^2} = 1 + {\left( {\cot A} \right)^2}$, we can

$ \Rightarrow {\text{cosec}}A = \sqrt {1 + {{\left( {\cot A} \right)}^2}} $

Using the above equation, equation (1) becomes

$ \Rightarrow \sin A = \dfrac{1}{{\sqrt {1 + {{\left( {\cot A} \right)}^2}} }}$

This above equation represents the sine trigonometric function in terms of cotangent trigonometric function.

Since, tangent trigonometric function is the reciprocal of cotangent trigonometric function.

Also we know that ${\left( {\sec A} \right)^2} = 1 + {\left( {\tan A} \right)^2}{\text{ }} \to {\text{(2)}}$ and $\tan A = \dfrac{1}{{\cot A}}{\text{ }} \to {\text{(3)}}$

Substituting the value of tangent trigonometric function from equation (3) in equation (2), we get

\[

{\left( {\sec A} \right)^2} = 1 + {\left( {\dfrac{1}{{\cot A}}{\text{ }}} \right)^2} = 1 + \dfrac{1}{{{{\left( {\cot A} \right)}^2}}} \\

\Rightarrow \sec A = \sqrt {1 + \dfrac{1}{{{{\left( {\cot A} \right)}^2}}}} \\

\]

This above equation represents the secant trigonometric function in terms of cotangent trigonometric function.

According to equation (3), we can write

$\tan A = \dfrac{1}{{\cot A}}$

This above equation represents the tangent trigonometric function in terms of cotangent trigonometric function.

Hence, $\sin A = \dfrac{1}{{\sqrt {1 + {{\left( {\cot A} \right)}^2}} }}$, \[\sec A = \sqrt {1 + \dfrac{1}{{{{\left( {\cot A} \right)}^2}}}} \] and $\tan A = \dfrac{1}{{\cot A}}$

Note: In this particular problem, in order to represent $\sin A$ in terms of $\cot A$ we will firstly convert $\sin A$ in terms of ${\text{cosec}}A$ and then finally we will convert ${\text{cosec}}A$ in terms of $\cot A$. Also, in order to convert $\sec A$ in terms of $\cot A$ we will firstly convert $\sec A$ in terms of $\tan A$ and then finally we will convert $\tan A$ in terms of $\cot A$. Also, for representing $\tan A$ in terms of $\cot A$ we will simply use the formula $\tan A = \dfrac{1}{{\cot A}}$.

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