Question

If $3cot{\rm{A}} = 4$, check whether $\dfrac{{1 - {{\tan }^2}{\rm{A}}}}{{1 + {{\tan }^2}{\rm{A}}}} = {\cos ^2}{\rm{A}} - {\sin ^2}{\rm{A}}$ or not.

Answer 1

We can use the right-angle triangle Pythagoras theorem and then find the values of $\dfrac{{1 - {{\tan }^2}{\rm{A}}}}{{1 + {{\tan }^2}{\rm{A}}}}$ and ${\cos ^2}{\rm{A}} - {\sin ^2}{\rm{A}}$. Right angle means the angle is $90^\circ$. It consists of six trigonometric ratios such as sin, cosine, tan, cot, sec and cosec.

Complete step by step solution:
The given trigonometric equation is $3cot{\rm{A}} = 4$.
Let us first rearrange the above trigonometric equation:
$cot{\rm{A}} = \dfrac{4}{3}$
We know the formula $\tan {\rm{A}} = \dfrac{1}{{\cot {\rm{A}}}}$ and substitute the value $cot{\rm{A}} = \dfrac{4}{3}$ in $\tan {\rm{A}} = \dfrac{1}{{\cot {\rm{A}}}}$.
$\tan {\rm{A}} = \dfrac{1}{{\dfrac{4}{3}}}\\\ = \dfrac{3}{4}$
Hence, the value of $\tan {\rm{A}}$ through the above result is $\dfrac{3}{4}$.
Now we can now draw a right-angled triangle on the basis of $\tan {\rm{A}} = \dfrac{3}{4}$.
Now, apply the right-angled triangle Pythagoras theorem for the above triangle.
${\left( {{\rm{Hypotenuse}}} \right)^2} = {\left( {{\rm{Height}}} \right)^2} + {\left( {{\rm{Base}}} \right)^2}\\\ {\rm{A}}{{\rm{C}}^2} = {\rm{A}}{{\rm{B}}^2} + {\rm{B}}{{\rm{C}}^{\rm{2}}}$
Substitute the values from the above diagram ${\rm{AB}} = 4$ and ${\rm{ BC}} = 3$ in ${\rm{A}}{{\rm{C}}^2} = {\rm{A}}{{\rm{B}}^2} + {\rm{B}}{{\rm{C}}^{\rm{2}}}$.
${\rm{A}}{{\rm{C}}^2} = {\rm{A}}{{\rm{B}}^2} + {\rm{B}}{{\rm{C}}^{\rm{2}}}\\\ = {\left( 4 \right)^2} + {\left( 3 \right)^2}\\\ = 16 + 9\\\ = 25$
Take the square root of the above equation:

${\rm{AC}} = \sqrt {25} \\\ = 5$
Hence, the hypotenuse of the triangle from the above result is 5.
Now we can use the basic formula of sine to calculate the value of $\sin {\rm{A}}$.
$\sin {\rm{A}} = \dfrac{{{\rm{Base}}}}{{{\rm{Hypotenuse}}}}\\\ = \dfrac{{{\rm{BC}}}}{{{\rm{AC}}}}\\\ = \dfrac{3}{5}$
Hence, the value of $\sin {\rm{A}}$ from the above result is $\dfrac{3}{5}$.
Now, we can use the basic formula of sine to calculate the value of $\cos {\rm{A}}$.
$\cos {\rm{A}} = \dfrac{{{\rm{Height}}}}{{{\rm{Hypotenuse}}}}\\\ = \dfrac{{{\rm{AB}}}}{{{\rm{AC}}}}\\\ = \dfrac{4}{5}$
Hence, the value of $\cos {\rm{A}}$ from the above result is $\dfrac{4}{5}$.
Now, we have to check whether,
$\dfrac{{1 - {{\tan }^2}{\rm{A}}}}{{1 + {{\tan }^2}{\rm{A}}}} = {\cos ^2}{\rm{A}} - {\sin ^2}{\rm{A}}$
Take the left-hand side of the above equation and substitute the value $\tan {\rm{A}} = \dfrac{3}{4}$ in $\dfrac{{1 - {{\tan }^2}{\rm{A}}}}{{1 + {{\tan }^2}{\rm{A}}}}$.
$\dfrac{{1 - {{\tan }^2}{\rm{A}}}}{{1 + {{\tan }^2}{\rm{A}}}} = \dfrac{{1 - {{\left( {\dfrac{3}{4}} \right)}^2}}}{{1 + {{\left( {\dfrac{3}{4}} \right)}^2}}}\\\ = \dfrac{{1 - \dfrac{9}{{16}}}}{{1 + \dfrac{9}{{16}}}}\\\ = \dfrac{{\dfrac{7}{{16}}}}{{\dfrac{{25}}{{16}}}}\\\ = \dfrac{7}{{25}}$
Now, take the right-hand side of the equation ${\cos ^2}{\rm{A}} - {\sin ^2}{\rm{A}}$ and substitute the values $\cos {\rm{A}} = \dfrac{4}{5}{\rm{ and }}\sin {\rm{A}} = \dfrac{3}{5}$ in ${\cos ^2}{\rm{A}} - {\sin ^2}{\rm{A}}$.
${\cos ^2}{\rm{A}} - {\sin ^2}{\rm{A}} = {\left( {\dfrac{4}{5}} \right)^2} - {\left( {\dfrac{3}{5}} \right)^2}\\\ = \dfrac{{16}}{{25}} - \dfrac{9}{{25}}\\\ = \dfrac{7}{{25}}$
Hence, the left hand side is the same as right hand side.

Note: Here we use the Pythagoras theorem. The trigonometric values cosec, sec and cot are the opposite values of sine, cos and tan respectively. We solve the question with the left-hand side and right-hand side process.