Question

If \[\dfrac{{\cos \alpha }}{{\cos \beta }} = m\] and \[\dfrac{{\cos \alpha }}{{\sin \beta }} = n\] , then show that, \[\left( {{m^2} + {n^2}} \right){\cos ^2}\beta = {n^2}\] ?

Answer 1

Hint: We can prove the given statement by using the basic trigonometric identities. and the basic idea to prove the given statement is that first we have to take the LHS part of the given equation then we have to substitute the values of m and on simplification we get the required RHS value.

Complete step by step solution:
Now let us consider the given data in the problem
 \[m = \dfrac{{\cos \alpha }}{{\cos \beta }} - - - - \left( 1 \right)\]
 \[n = \dfrac{{\cos \alpha }}{{\sin \beta }} - - - - \left( 2 \right)\]
After squaring m and n we will have
 \[{m^2} = \dfrac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\beta }}\]
and
 \[{n^2} = \dfrac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\beta }}\]
Now consider only LHS part of the given equation which has to be proved
 \[LHS = \left( {{m^2} + {n^2}} \right){\cos ^2}\beta \]
On substituting the values of m and n from equations 1 and 2 we get
 \[LHS = \left( {\dfrac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\beta }} + \dfrac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\beta }}} \right){\cos ^2}\beta \]
Taking the LCM of the denominator we get
 \[LHS = \left( {\dfrac{{{{\cos }^2}\alpha {{\sin }^2}\beta + {{\cos }^2}\alpha {{\cos }^2}\beta }}{{{{\cos }^2}\beta {{\sin }^2}\beta }}} \right){\cos ^2}\beta \]
Now taking \[{\cos ^2}\alpha \] as common term in numerator we get
 \[LHS = {\cos ^2}\alpha \left( {\dfrac{{{{\sin }^2}\beta + {{\cos }^2}\beta }}{{{{\cos }^2}\beta {{\sin }^2}\beta }}} \right){\cos ^2}\beta \]
Now use the trigonometric identity \[{\sin ^2}\beta + {\cos ^2}\beta = 1\] we get
 \[LHS = {\cos ^2}\alpha \left( {\dfrac{1}{{{{\cos }^2}\beta {{\sin }^2}\beta }}} \right){\cos ^2}\beta \]
After simplification terms get canceled then we get
 \[LHS = \left( {\dfrac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\beta }}} \right)\]
 \[LHS = {\left( {\dfrac{{\cos \alpha }}{{\sin \beta }}} \right)^2}\]
from equation (2) we have
 \[LHS = {\left( n \right)^2}\]
 \[LHS = RHS\]
Hence proved

Note: Trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation. Identity inequalities which are true for every value occurring on both sides of an equation. Geometrically, these identities involve certain functions of one or more angles. There are various distinct identities involving the side length as well as the angle of a triangle. The trigonometric identities hold true only for the right-angle triangle.