If a \[a{{\sin }^{2}}\alpha +b{{\cos }^{2}}\alpha =p,b{{\sin }^{2}}\beta +a{{\cos }^{2}}\beta =q,a\tan \alpha =b\tan \beta \], Show that \[\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{p}+\dfrac{1}{q}\], where \[a\ne p\] and all of them are non-zero.

Hint: Given three equations,divide First equation by \[{{\cos }^{2}}\alpha \] and second equation by \[{{\cos }^{2}}\beta \].Then substitute the values in third equation and simplify it.

“Complete step-by-step answer:”

Given that \[a{{\sin }^{2}}\alpha +b{{\cos }^{2}}\alpha =p-(1)\]
Now, divide both sides by \[{{\cos }^{2}}\alpha \].
\[\dfrac{a{{\sin }^{2}}\alpha +b{{\cos }^{2}}\alpha }{{{\cos }^{2}}\alpha }=\dfrac{p}{{{\cos }^{2}}\alpha }\]
\[\because \]We know that \[\dfrac{\sin \alpha }{\cos \alpha }=\tan \alpha \]
\[\dfrac{1}{\cos \alpha }=\sec \alpha \]
\[\begin{align}
  & a{{\tan }^{2}}\alpha +b=p{{\sec }^{2}}\alpha \\
 & \because {{\sec }^{2}}\alpha =1+{{\tan }^{2}}\alpha \\
 & \Rightarrow a{{\tan }^{2}}\alpha +b=p\left( 1+{{\tan }^{2}}\alpha \right) \\
 & a{{\tan }^{2}}\alpha +b=p+p{{\tan }^{2}}\alpha \\
 & a{{\tan }^{2}}\alpha -p{{\tan }^{2}}\alpha =p-b \\
 & {{\tan }^{2}}\alpha \left( a-p \right)=p-b \\
 & \Rightarrow {{\tan }^{2}}\alpha =\dfrac{p-b}{a-p}-\left( 2 \right) \\
\end{align}\]
Given that \[b{{\sin }^{2}}\beta +a{{\cos }^{2}}\beta =q-(3)\]
Now, divide both sides by\[{{\cos }^{2}}\beta \].
\[\begin{align}
  & \dfrac{b{{\sin }^{2}}\beta +a{{\cos }^{2}}\beta }{{{\cos }^{2}}\beta }=\dfrac{q}{{{\cos }^{2}}\beta } \\
 & b{{\tan }^{2}}\beta +a=q{{\sec }^{2}}\beta \\
 & b{{\tan }^{2}}\beta +a=q\left( 1+{{\tan }^{2}}\beta \right) \\
 & b{{\tan }^{2}}\beta +a=q+q{{\tan }^{2}}\beta \\
 & b{{\tan }^{2}}\beta -q{{\tan }^{2}}\beta =q-a \\
 & {{\tan }^{2}}\beta \left( b-q \right)=q-a \\
 & \therefore {{\tan }^{2}}\beta =\dfrac{q-a}{b-q}-(4) \\
\end{align}\]
From the question, \[a\tan \alpha =b\tan \beta \].
Squaring on both sides we get,
\[\begin{align}
  & {{\left( a\tan \alpha \right)}^{2}}={{\left( b\tan \beta \right)}^{2}} \\
 & {{a}^{2}}{{\tan }^{2}}\alpha ={{b}^{2}}{{\tan }^{2}}\beta \\
 & \Rightarrow \dfrac{{{\tan }^{2}}\alpha }{{{\tan }^{2}}\beta }=\dfrac{{{b}^{2}}}{{{a}^{2}}}-\left( 5 \right) \\
\end{align}\]
From (3) and (4) substitute the values of (3) and (4) in (5).
\[\begin{align}
  & \dfrac{\dfrac{\left( p-b \right)}{\left( a-p \right)}}{\dfrac{\left( q-a \right)}{\left( b-q \right)}}=\dfrac{{{b}^{2}}}{{{a}^{2}}} \\
 & \Rightarrow \dfrac{\left( p-b \right)\left( b-q \right)}{\left( a-p \right)\left( q-a \right)}=\dfrac{{{b}^{2}}}{{{a}^{2}}} \\
 & {{a}^{2}}\left[ \left( p-b \right)\left( b-q \right) \right]={{b}^{2}}\left[ \left( a-p \right)\left( q-a \right) \right] \\
\end{align}\]
Opening the brackets and simplifying it,
\[\begin{align}
  & {{a}^{2}}\left[ pb-pq-{{b}^{2}}+bq \right]={{b}^{2}}\left[ aq-{{a}^{2}}-pq+ap \right] \\
 & \Rightarrow {{a}^{2}}pb-{{a}^{2}}pq-{{a}^{2}}{{b}^{2}}+{{a}^{2}}bq=a{{b}^{2}}q-{{a}^{2}}{{b}^{2}}-{{b}^{2}}pq+a{{b}^{2}}q \\
\end{align}\]
Cancel out \[{{a}^{2}}{{b}^{2}}\] on both sides.
\[\begin{align}
  & {{a}^{2}}pb-{{a}^{2}}pq+{{a}^{2}}bq-a{{b}^{2}}q+{{b}^{2}}pq-a{{b}^{2}}p=0 \\
 & \left( {{a}^{2}}pb-a{{b}^{2}}p \right)-\left( {{a}^{2}}pq-{{b}^{2}}pq \right)+q\left( {{a}^{2}}b-a{{b}^{2}} \right)=0 \\
 & \Rightarrow abp\left( a-b \right)-pq\left( {{a}^{2}}-{{b}^{2}} \right)+abq\left( a-b \right)=0 \\
\end{align}\]
We know, \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\]
\[\begin{align}
  & \Rightarrow abp-pq\left( a+b \right)+abq=0 \\
 & abp+abq=pq\left( a+b \right) \\
 & ab\left( p+q \right)=pq\left( a+b \right) \\
 & \Rightarrow \dfrac{p+q}{pq}=\dfrac{a+b}{ab} \\
\end{align}\]
\[\Rightarrow \]By dividing and simplifying it,
\[\dfrac{1}{p}+\dfrac{1}{q}=\dfrac{1}{q}+\dfrac{1}{b}\]
Hence, proved.
Note: From \[a\tan \alpha =b\tan \beta \], solve them to find \[\dfrac{{{\tan }^{2}}\alpha }{{{\tan }^{2}}\beta }\]. By substituting the expression we get \[\dfrac{1}{p}+\dfrac{1}{q}=\dfrac{1}{q}+\dfrac{1}{b}\].