Question

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.

Answer 1

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 $\dfrac{\sin \alpha }{\cos \alpha }=\tan \alpha$
$\dfrac{1}{\cos \alpha }=\sec \alpha$

& 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{aligned}$$ Given that $$b{{\sin }^{2}}\beta +a{{\cos }^{2}}\beta =q-(3)$$ Now, divide both sides by$${{\cos }^{2}}\beta $$. $$\begin{aligned} & \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{aligned}$$ From the question, $$a\tan \alpha =b\tan \beta $$. Squaring on both sides we get, $$\begin{aligned} & {{\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{aligned}$$ From (3) and (4) substitute the values of (3) and (4) in (5). $$\begin{aligned} & \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{aligned}$$ Opening the brackets and simplifying it, $$\begin{aligned} & {{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{aligned}$$ Cancel out $${{a}^{2}}{{b}^{2}}$$ on both sides. $$\begin{aligned} & {{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{aligned}$$ We know, $${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$$ $$\begin{aligned} & \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{aligned}$$ $$\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}$$.