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Question: If \[\dfrac{{\cos \alpha }}{{\cos \beta }} = m\] and \[\dfrac{{\cos \alpha }}{{\sin \beta }} = n\] ,...

If cos⁡αcos⁡β=m\dfrac{{\cos \alpha }}{{\cos \beta }} = m and cos⁡αsin⁡β=n\dfrac{{\cos \alpha }}{{\sin \beta }} = n , then show that, (m2+n2)cos⁡2β=n2\left( {{m^2} + {n^2}} \right){\cos ^2}\beta = {n^2} ?

Explanation

Solution

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=cos⁡αcos⁡β−−−−(1)m = \dfrac{{\cos \alpha }}{{\cos \beta }} - - - - \left( 1 \right)
n=cos⁡αsin⁡β−−−−(2)n = \dfrac{{\cos \alpha }}{{\sin \beta }} - - - - \left( 2 \right)
After squaring m and n we will have
m2=cos⁡2αcos⁡2β{m^2} = \dfrac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\beta }}
and
n2=cos⁡2αsin⁡2β{n^2} = \dfrac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\beta }}
Now consider only LHS part of the given equation which has to be proved
LHS=(m2+n2)cos⁡2βLHS = \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=(cos⁡2αcos⁡2β+cos⁡2αsin⁡2β)cos⁡2βLHS = \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=(cos⁡2αsin⁡2β+cos⁡2αcos⁡2βcos⁡2βsin⁡2β)cos⁡2βLHS = \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⁡2α{\cos ^2}\alpha as common term in numerator we get
LHS=cos⁡2α(sin⁡2β+cos⁡2βcos⁡2βsin⁡2β)cos⁡2βLHS = {\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⁡2β+cos⁡2β=1{\sin ^2}\beta + {\cos ^2}\beta = 1 we get
LHS=cos⁡2α(1cos⁡2βsin⁡2β)cos⁡2βLHS = {\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=(cos⁡2αsin⁡2β)LHS = \left( {\dfrac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\beta }}} \right)
LHS=(cos⁡αsin⁡β)2LHS = {\left( {\dfrac{{\cos \alpha }}{{\sin \beta }}} \right)^2}
from equation (2) we have
LHS=(n)2LHS = {\left( n \right)^2}
LHS=RHSLHS = 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.