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)

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

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

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

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

  
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