Ex.8.4 Q2 Introduction to Trigonometry Solution - NCERT Maths Class 10

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Question

Write all the other trigonometric ratios of\(\begin{align}\angle \rm{A}\end{align}\) in terms of \(sec \;A\).

Text Solution

 

Reasoning:

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

Steps:

We know that,

Trigonometric Function,  \(\begin{align}\text{cos}\,{A = }\frac{1}{\sec {A}}\ldots \text{Equation }\left( \text{1} \right)\end{align}\)

Also,

\[\begin{align}\text{si}{{\text{n}}^{\text{2}}}A + \text{co}{{\text{s}}^{\text{2}}}{A} &= {1 }\left( \text{Trigonometric identity} \right) \\ \text{si}{{\text{n}}^{\text{2}}}{A} &= {1 }-\text{ co}{{\text{s}}^{\text{2}}}{A }\left( \text{By transposing} \right) \\ \end{align}\]

Using value of \(\text{cos} \;A\) from Equation (1) and simplifying further,

\[\begin{align}\sin {{A} }&=\sqrt {1 - {{\left( {\frac{1}{{\sec {{A}}}}} \right)}^2}} \\&= \sqrt {\frac{{{{\sec }^2}\,{{A}} - 1}}{{{{\sec }^2}\,{{A}}}}} \\ &= \frac{{\sqrt {{{\sec }^2}{{A}} - 1} }}{{\sec {{A}}}} & \ldots {\text{Equation (2)}} \end{align}\]

\[\begin{align}{\text{ta}}{{\text{n}}^{\text{2}}}{{A + 1 }}&={{\text{sec}}^{\text{2}}}{{A }}\left( {{\text{Trigonometric identity}}} \right)\\ {\text{ta}}{{\text{n}}^{\text{2}}}{{A}}&={{\text{sec}}^{\text{2}}}{{A - 1 }}\left( {{\text{By transposing}}} \right) \\\end{align}\]

Trigonometric Function,

\(\begin{align} {\text{tan}}\,{\text{A}}&=\sqrt {{\text{se}}{{\text{c}}^2}{\text{A}} - {\text{1}}} {\text{ }} ...{\text{ Equation }}\left( {\text{3}} \right)\\ {\text{cot}}\,{\text{A}}\,&=\,\frac{{{\text{cosA}}}}{{{\text{sinA}}}} \end{align}\)
\(\begin{align}= \frac{{\frac{1}{{\sec \,{\text{A}}}}}}{{\frac{{\sqrt {{{\sec }^2}} {\text{A}} - 1}}{{\sec \,{\text{A}}}}}} \ldots  \end{align}\) (By substituting equations (1) and (2) )
\(\begin{align}= \frac{1}{{\sqrt {{{\sec }^2}{\text{A}} - {\text{1}}} }}   \end{align}\)
\(\begin{align}{\rm{cosec}}\,{\text{A}}=\frac{1}{{\sin {\text{A}}}}  \end{align}\)
\(\begin{align}  = \frac{{\sec {\text{A}}}}{{\sqrt {{{\sec }^2}\,{\text{A}} - 1} }}  \end{align}\)  (By substituting Equation (2) and simplifying)

  
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