Question

If $\dfrac{{\cos \alpha }}{{\sin \beta }} = n$ and $\dfrac{{\cos \alpha }}{{\cos \beta }} = m$ then find the value of ${\cos ^2}\beta $ is

A. $\dfrac{1}{{{m^2} + {n^2}}}$

B. $\dfrac{{{m^2}}}{{{m^2} + {n^2}}}$

C. $\dfrac{{{n^2}}}{{{m^2} + {n^2}}}$

D. $0$

Answer 1

Hint: In order to solve the trigonometry function we have to first see what is the requirement of the question and according to that we have to apply some arithmetic operations to get the final value. We can do division, square on the equation to get the required answer.

Complete step-by-step answer:
Given that $\dfrac{{\cos \alpha }}{{\sin \beta }} = n$ and $\dfrac{{\cos \alpha }}{{\cos \beta }} = m$

Let us assume
$\dfrac{{\cos \alpha }}{{\sin \beta }} = n \to (1)$
$\dfrac{{\cos \alpha }}{{\cos \beta }} = m \to (2)$
When we do equation $\dfrac{{(2)}}{{(1)}}$ we get
$
  \dfrac{{\dfrac{{\cos \alpha }}{{\cos \beta }}}}{{\dfrac{{\cos \alpha }}{{\sin \beta }}}} = \dfrac{m}{n} \\
   \Rightarrow \dfrac{{\cos \alpha }}{{\cos \beta }} \times \dfrac{{\sin \beta }}{{\cos \alpha }} = \dfrac{m}{n} \\
   \Rightarrow \dfrac{{\sin \beta }}{{\cos \beta }} = \dfrac{m}{n} \\
$

By doing square both side we get

$
   \Rightarrow \dfrac{{\sin \beta }}{{\cos \beta }} = \dfrac{m}{n} \\
   \Rightarrow {\left( {\dfrac{{\sin \beta }}{{\cos \beta }}} \right)^2} = {\left( {\dfrac{m}{n}} \right)^2} \\
   \Rightarrow \left( {\dfrac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\beta }}} \right) = \left( {\dfrac{{{m^2}}}{{{n^2}}}} \right) \\
$

By adding \[1\] both side we get
$
    \\
   \Rightarrow \left( {\dfrac{{{{\sin }^2}\beta }}{{{{\cos }^2}\beta }}} \right) + 1 = \left( {\dfrac{{{m^2}}}{{{n^2}}}} \right) + 1 \\
   \Rightarrow \left( {\dfrac{{{{\sin }^2}\beta + {{\cos }^2}\beta }}{{{{\cos }^2}\beta }}} \right) = \left( {\dfrac{{{m^2} + {n^2}}}{{{n^2}}}} \right) \\
  \because ({\sin ^2}\beta + {\cos ^2}\beta = 1) \\
   \Rightarrow \left( {\dfrac{1}{{{{\cos }^2}\beta }}} \right) = \left( {\dfrac{{{m^2} + {n^2}}}{{{n^2}}}} \right) \\
$
By cross multiplication of the above equation we get
$
  {n^2}(1) = ({m^2} + {n^2})({\cos ^2}\beta ) \\
   \Rightarrow \left( {\dfrac{{{n^2}}}{{{m^2} + {n^2}}}} \right) = {\cos ^2}\beta \\
 $

So here we can see that we get
$
    \\
  {\cos ^2}\beta = \left( {\dfrac{{{n^2}}}{{{m^2} + {n^2}}}} \right) \\
$

Hence, \[{\cos ^2}\beta = \left( {\dfrac{{{n^2}}}{{{m^2} + {n^2}}}} \right)\] .

Therefore C is the correct option.


Note- For solving trigonometry functions we should know about their relationship. There are three main relations which can be used to solve trigonometry questions that is $({\sin ^2}\theta + {\cos ^2}\theta = 1),$ $({\sec ^2}\theta = 1 + {\tan ^2}\theta )$ and $(\cos e{c^2}\theta = 1 + {\cot ^2}\theta )$ We can also prove it by using basic trigonometric properties.