# Ex.8.4 Q1 Introduction to Trigonometry Solution - NCERT Maths Class 10

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## Question

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

## Text Solution

#### Reasoning:

\begin{align}{ \text{cosec} ^{2} A=1+\cot ^{2} A} \\ {\sec ^{2} A=1+\tan ^{2} A}\end{align}

#### Steps:

Consider a  \begin{align}\Delta {ABC}\end{align} with \begin{align}\angle {B }=\text{ }90{}^\circ \end{align}

Using the Trigonometric Identity,

$$\text{cose}{{\text{c}}^{2}}\,{A}=1+{{\cot }^{2}}{A}$$

\begin{align} \frac{1}{\text{cose}{{\text{c}}^{2}}\,{A}}&=\frac{\text{1}}{\text{1+co}{{\text{t}}^{\text{2}}}{A}} \end{align} (By taking reciprocal both the sides)

\begin{align} &{{\text{sin}}^{2}}{A}=\frac{1}{1+{{\cot }^{2}}{A}} \\ & \left( \text{As }\frac{1}{\text{cose}{{\text{c}}^{2}}\,\text{A}}={{\sin }^{2}}\text{A} \right) \\ \end{align}

Therefore,

\begin{align}\text{sin}\,{A}\,=+\frac{1}{\sqrt{1+{{\cot }^{2}}{A}}} \end{align}

For any sine value with respect to an angle in a triangle, sine value will never be negative. Since, sine value will be negative for all angles greater than 180°.

Therefore, \begin{align}\text{sin}\,{A}\,=\frac{\text{1}}{\sqrt{\text{1+co}{{\text{t}}^{\text{2}}}{A}}} \end{align}

We know that,

\begin{align}\text{tan} A=\frac{\text{sin}} A{\cos \,{A}}\end{align}

However, Trigonometric Function,

\begin{align}\text{cot}\,{A=}\frac{\cos \,{A}}{\text{Sin}\,{A}}\end{align}

Therefore, Trigonometric Function,

\begin{align}\text{tan}\,A=\frac{1}{\cot \,{A}}\end{align}

Also,

\begin{align}\text{se}{{\text{c}}^{\text{2}}} A = 1 + {{\text{tan}}^{\text{2}}}{A}\end{align} (Trigonometric Identity)

\begin{align}=1+\frac{1}{{{\cos }^{2}}{A}} \\ =\frac{{{\cot }^{2}}\,{A+1}}{{{\cot }^{2}}{A}} \\ \end{align}

\begin{align}\text{sec}\,{A}=\frac{\sqrt{{{\cot }^{2}}{A+1}}}{\cot \,A}\end{align}

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