     Question Answers

# 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.

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 $\dfrac{1}{p}+\dfrac{1}{q}=\dfrac{1}{q}+\dfrac{1}{b}$.

View Notes
Trigonometric Equations  CBSE Class 12 Maths Formulas  CBSE Class 12 Maths Chapter-9 Differential Equations Formula  CBSE Class 12 Maths Chapter-2 Inverse Trigonometric Functions Formula  Differential Equations For Class 12  CBSE Class 12 Maths Chapter-12 Linear Programming Formula  CBSE Class 12 Maths Chapter-7 Integrals Formula  CBSE Class 12 Maths Chapter-13 Probability Formula  CBSE Class 12 Maths Chapter-4 Determinants Formula  CBSE Class 12 Maths Chapter-3 Matrices Formula  