Ex.8.1 Q8 Introduction to Trigonometry Solution - NCERT Maths Class 10

Question

If $$3 \;cot A = 4$$, check whether \begin{align} \frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = {\cos ^2}A - {\sin ^2}A \end{align} or not.

Video Solution
Introduction To Trigonometry
Ex 8.1 | Question 8

Text Solution

What is known?

Cotangent of angle $$A$$

What is unknown?

Whether \begin{align} \frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = {\cos ^2}A - {\sin ^2}A \end{align}

Reasoning:

Using $$3\cot {{A}} = 4$$, 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:

\begin{align}{\rm{3}}\,{\rm{cot}}\,{{A}}\,\, &= \,\,{{4}}\\{\rm{cot}}\,{{A}}\,\, &= \,\,\frac{{\rm{4}}}{{\rm{3}}}\end{align}

Let , in which angle $$B$$ is right angle.

\begin{align} \cot A & = \frac{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\angle A}}{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\angle A}} \\ & = \frac{{AB}}{{BC}} = \frac{4}{3} \end{align}

Let $$AB = 4k\;{\rm{ }}{\rm{and }}\;BC = 3k$$ where $$k$$ is a positive integer.

By applying Pythagoras theorem in $$\Delta ABC$$ we get.

\begin{align}{A}{C^2}\, & = {A}{{B}^2} + {B}{C^2}\\ &= {(4k)^2} + {(3k)^2}\\ &= 16{k^2} + 9{k^2}\\ &= 25{k^2}\\{{AC}}\,& = \,\sqrt {{{25k}}{\,^{{2}}}} \\&={{5k}}\end{align}

Therefore,

\begin{align} {\rm{tan}}\,A &= \frac{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\angle A}}{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\angle A}}\\ & = \frac{{BC}}{{AB}} \\ & = \frac{{3k}}{{4k}} \\ &= \frac{3}{4} \end{align}

\begin{align}{\text{sin}} A &= \frac{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\angle {{A}}}}{{{\rm{hypotenuse}}}}{\rm{ = }}\frac{{{{BC}}}}{{{{AC}}}}\\&=\frac{{{\rm{3k}}}}{{{\rm{5k}}}}= \frac{{\rm{3}}}{{\rm{5}}} \end{align}

\begin{align} \cos A & = \frac{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\angle A}}{{{\rm{hypotenuse}}}} \\ & = \frac{{AB}}{{AC}} \\ & = \frac{{4k}}{{5k}} \\ & = \frac{4}{5} \end{align}

\begin{align}L.H.S &= \frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}}\\ &= \frac{{1 - {{\left( {\frac{3}{4}} \right)}^2}}}{{1 + {{\left( {\frac{3}{4}} \right)}^2}}}\\ &= \frac{{1 - \frac{9}{{16}}}}{{1 + \frac{9}{{16}}}}\\& = \frac{{16 - 9}}{{16 + 9}}\\ &= \frac{7}{{25}}\end{align}

\begin{align}R.H.S\,\, &= {\cos ^2}A - {\sin ^2}A\\&= {\left( {\frac{4}{5}} \right)^2} - {\left( {\frac{3}{5}} \right)^2}\\ &= \frac{{16}}{{25}} - \frac{9}{{25}}\\&= \frac{{16 - 9}}{{25}}\\& = \frac{7}{{25}}\end{align}

Therefore, \begin{align}\frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = {\cos ^2}A - {\sin ^2}A \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