Question

If $3\cot \left(A\right)=4$, check whether$\frac{(1-{\tan }^{2}A)}{(1+{\tan }^{2}A)}={\cos }^{2}A–{\sin }^{2}A$ or not.

Answer 1

Step 1. Find the value of trigonometric ratios

It is given that $3cotA=4$

Let $△ABC$ in which$\angle B=90°$

We know that the $\cot$ function is the reciprocal of the tan function and it is written as

$cot\left(A\right)=\frac{AB}{BC}=\frac{4}{3}$

Let us consider $AB=4k,BC=3k$

where $k$ is a positive real number.

According to the Pythagorean theorem,

$$\begin{gathered} A{C}^2=A{B}^2+B{C}^2 \\ \Rightarrow A{C}^2={\left(4k\right)}^2+{\left(3k\right)}^2 \\ \Rightarrow A{C}^2=16k^2+9k^2 \\ \Rightarrow A{C}^2=25K^2 \\ \Rightarrow AC=5k \end{gathered}$$

Now, apply the values corresponding to the ratios

$$\begin{gathered} \Rightarrow \tan A=\frac{BC}{AB}=\frac{3}{4} \\ \Rightarrow \sin A=\frac{BC}{AC}=\frac{3}{5} \\ \Rightarrow \cos A=\frac{AB}{AC}=\frac{4}{5} \end{gathered}$$

Step 2. Prove the given expression

Now compare the left-hand side(LHS) with right-hand side(RHS).

First take Left hand side

$\begin{array}{rcl}\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}& =& \frac{1-{\left({\displaystyle \frac{3}{4}}\right)}^2}{1+{\left({\displaystyle \frac{3}{4}}\right)}^2}\\ & =& \frac{1-{\displaystyle \frac{9}{16}}}{1+{\displaystyle \frac{9}{16}}}\\ & =& \frac{7}{25}\end{array}$

Now take the Right hand side

$\begin{array}{rcl}{\cos }^{2}A-{\sin }^{2}A& =& {\left(\frac{4}{5}\right)}^2-{\left(\frac{3}{5}\right)}^2\\ & =& \frac{16}{25}-\frac{9}{25}\\ & =& \frac{7}{25}\end{array}$

Therefore ,Left hand side equals to Right hand side.

Hence, $\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}={\cos }^{2}A-{\sin }^{2}A$ is proved