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

Go back to  'Ex.8.1'

## Question

If $$\cot \theta = \frac{7}{8}$$, evaluate:

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

(ii) \begin{align} {\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 &= \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 }} \\ & \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}

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school