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

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Question

Q8. 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.

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.

\[\cot A = \frac{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\angle A}}{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\angle A}} = \frac{{AB}}{{BC}} = \frac{4}{3}\]

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,

\[{\rm{tan}}\,A = \frac{{{\rm{side}}\;{\rm{opposite}}\;{\rm{to}}\;\angle A}}{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\angle A}} = \frac{{BC}}{{AB}} = \frac{{3k}}{{4k}} = \frac{3}{4}\]

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

\[\cos A = \frac{{{\rm{side}}\;{\rm{adjacent}}\;{\rm{to}}\;\angle A}}{{{\rm{hypotenuse}}}} = \frac{{AB}}{{AC}} = \frac{{4k}}{{5k}} = \frac{4}{5}\]

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

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

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