Ex.8.1 Q7 Introduction to Trigonometry Solution - NCERT Maths Class 10

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Question

 If \(\cot \theta  = \frac{7}{8}\), evaluate: \(\begin{align}\rm( i )\,\,{\rm{ }}\frac{{\left( {1 + \sin \theta } \right)\left( {1 - \sin \theta } \right)}}{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}}\end{align}\),    \(\begin{align}\left( \rm{ii} \right){\rm{  co}}{{\rm{t}}^2}\theta \end{align} \)

Text Solution

What is the known?

\(\cot \theta  = \frac{7}{8}\)

What is the unknown?

Value of \(\begin{align} \left( \rm{i} \right)\,\,{\rm{  }}\frac{{\left( {1 + \sin \theta } \right)\left( {1 - \sin \theta } \right)}}{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}} \end{align}\), and   \(\begin{align} \left( \rm{ii} \right){\rm{  co}}{{\rm{t}}^2}\theta \end{align} \)

Reasoning:

Using \(\cot \theta  = \frac{7}{8}\), we can find the ratio of the length of two sides of the right-angled triangle. Then by using Pythagoras theorem, the third side and required trigonometric ratios.

Steps:

Let \({\rm{\Delta }}\,{{ABC}}\), in which angle \(B\) is right angle.

\[\begin{align} \cot \theta  &= \frac{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\theta }}{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\theta }}\\ &= \frac{{AB}}{{BC}}\\&= \frac{7}{8} \end{align}\]

Let \(AB = 7k\)and \(BC = 8k\), where \(k\) is a positive integer.

By applying Pythagoras theorem in we get.

\[\begin{align}{\rm A}{C^2}\, & = {\rm A}{{\rm B}^2} + {\rm B}{C^2}\\ &= {(7k)^2} + {(8k)^2}\\ &= 49{k^2} + 64{k^2}\\ &= 113{k^2}\\
AC\, &= \,\sqrt {113k{\,^2}} \\ &= \,\sqrt {113} k\end{align}\]

Therefore,

\[\begin{align}{\rm{sin }}\theta {\rm{ = }}\frac{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\theta }}{{{\rm{hypotenuse}}}} = \frac{{BC}}{{AC}} = \frac{{8k}}{{\sqrt {113} k}} = \frac{8}{{\sqrt {113} }}\\\cos \theta  = \frac{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\theta }}{{{\rm{hypotenuse}}}} = \frac{{AB}}{{AC}} = \frac{{7k}}{{\sqrt {113} k}} = \frac{7}{{\sqrt {113} }}\end{align}\]

\(\begin{align}\left(\rm{i} \right)\,\,{\rm{  }}\frac{{\left( {1 + \sin \theta } \right)\left( {1 - \sin \theta } \right)}}{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}} \end{align}\)

\[\begin{align}\frac{{\left( {1 + \sin \theta } \right)\left( {1 - \sin \theta } \right)}}{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}} &= \frac{{1 - {{\sin }^2}\theta }}{{1 - {{\cos }^2}\theta }}\qquad\left[ \because {\left( {a + b} \right)\left( {a - b} \right) = \left( {{a^2} - {b^2}} \right)} \right]\\ &= \frac{{1 - {{\left( {\frac{8}{{\sqrt {113} }}} \right)}^2}}}{{1 - {{\left( {\frac{7}{{\sqrt {113} }}} \right)}^2}}}\\ &= \frac{{1 - \frac{{64}}{{113}}}}{{1 - \frac{{49}}{{113}}}}\\
 &= \frac{{\frac{{49}}{{113}}}}{{\frac{{64}}{{113}}}}\\ &= \frac{{49}}{{64}}\end{align}\]

\(\left( \rm{ii} \right){\rm{  co}}{{\rm{t}}^2}\theta \)

\[\begin{align}{\rm{co}}{{\rm{t}}^2}\theta  &= {\left( {\frac{7}{8}} \right)^2}\\ &= \frac{{49}}{{64}}\end{align}\]