Question

Prove: $\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A$ .

Answer 1

Hint: Useful Trigonometric identities. With the help of given identities we will be able to prove the given statement.
 ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
 ${{\tan }^{2}}\theta +1={{\sec }^{2}}\theta $
 $1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta $
 $\tan \theta =\dfrac{sin\theta }{\cos \theta }$
 $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$
 $\csc \theta =\dfrac{1}{\sin \theta }$
 $\sec \theta =\dfrac{1}{\cos \theta }$
 $\cot \theta =\dfrac{1}{\tan \theta }$

Complete step-by-step answer:
Let's consider each of the expressions one by one:
LHS = $\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)$
Using the identities ${{\tan }^{2}}\theta +1={{\sec }^{2}}\theta $ and $1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta $ , we get:
 $\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\sec }^{2}}A}{{{\csc }^{2}}A}$
We know that $\sec \theta =\dfrac{1}{\cos \theta }$ and $\csc \theta =\dfrac{1}{\sin \theta }$ , therefore:
$\Rightarrow$ $\dfrac{{{\sec }^{2}}A}{{{\csc }^{2}}A}=\dfrac{\left( \dfrac{1}{{{\cos }^{2}}A} \right)}{\left( \dfrac{1}{{{\sin }^{2}}A} \right)}$
In order to divide by a fraction, we have to multiply with the fractions reciprocal:
$\Rightarrow$ $\dfrac{\left( \dfrac{1}{{{\cos }^{2}}A} \right)}{\left( \dfrac{1}{{{\sin }^{2}}A} \right)}=\left( \dfrac{1}{{{\cos }^{2}}A} \right)\times \left( \dfrac{{{\sin }^{2}}A}{1} \right)=\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}={{\tan }^{2}}A$ . Hence proved.
And, RHS = ${{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$
Using $\tan \theta =\dfrac{sin\theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ , we can write it as:
 ${{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\left( \dfrac{1-\dfrac{\sin A}{\cos A}}{1-\dfrac{\cos A}{\sin A}} \right)}^{2}}$
On equating the denominators and subtracting, we have:
 $\Rightarrow$ ${{\left( \dfrac{1-\dfrac{\sin A}{\cos A}}{1-\dfrac{\cos A}{\sin A}} \right)}^{2}}={{\left( \dfrac{\dfrac{\cos A-\sin A}{\cos A}}{\dfrac{\sin A-\cos A}{\sin A}} \right)}^{2}}={{\left( \dfrac{\cos A-\sin A}{\cos A}\times \dfrac{\sin A}{\sin A-\cos A} \right)}^{2}}$
Since, $\tan \theta =\dfrac{sin\theta }{\cos \theta }$ and (cos A - sin A) = -(sin A - cos A), we can rewrite the expression as:
$\Rightarrow$ ${{\left[ \dfrac{\sin A}{\cos A}\times \dfrac{(-1)(\sin A-\cos A)}{(\sin A-\cos A)} \right]}^{2}}={{[\tan A\times (-1)]}^{2}}={{\tan }^{2}}A$ . Hence proved.

Note: In a right-angled triangle with length of the side opposite to angle θ as perpendicular (P), base (B) and hypotenuse (H):
 $\sin \theta =\dfrac{P}{H},\cos \theta =\dfrac{B}{H},\tan \theta =\dfrac{P}{B}$
 ${{P}^{2}}+{{B}^{2}}={{H}^{2}}$ (Pythagoras' Theorem)
The identities ${{\sin}^{2}}\theta +{{\cos }^{2}}\theta =1$ , ${{\tan }^{2}}\theta +1={{\sec }^{2}}\theta $ and $1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta $ are equivalent to each other and they are a direct result of the Pythagoras' theorem.