Question

$\dfrac{{\sin {\rm{A}}}}{{\sin \left( {90^\circ - {\rm{A}}} \right)}} + {\rm{\;}}\dfrac{{\cos {\rm{A}}}}{{\cos \left( {90^\circ - {\rm{A}}} \right)}} = {\rm{\;}}\sec \left( {90^\circ - {\rm{A}}} \right){\rm{cosec}}\left( {90^\circ - {\rm{A}}} \right)$

Answer 1

Hint: In this question we have to prove the given expression so we have to first simplify the LHS part and use the identities $\sin \left( {90^\circ - {\rm{A}}} \right) = \cos {\rm{A}}$ and $\cos \left( {90^\circ - {\rm{A}}} \right) = \sin {\rm{A}}$ and then make use of identity ${\sin ^2}{\rm{A}} + {\rm{\;}}{\cos ^2}{\rm{A}} = 1$ to get the RHS part.

Complete step-by-step answer:
We have to prove LHS = RHS
Now,
LHS = $\dfrac{{\sin {\rm{A}}}}{{\sin \left( {90^\circ - {\rm{A}}} \right)}} + {\rm{\;}}\dfrac{{\cos {\rm{A}}}}{{\cos \left( {90^\circ - {\rm{A}}} \right)}}$
As we know that, $\sin \left( {90^\circ - {\rm{A}}} \right) = \cos {\rm{A}}$ and $\cos \left( {90^\circ - {\rm{A}}} \right) = \sin {\rm{A}}$
LHS = $\dfrac{{\sin {\rm{A}}}}{{\cos {\rm{A}}}} + {\rm{\;}}\dfrac{{\cos {\rm{A}}}}{{\sin {\rm{A}}}}$
Taking LCM, we get
LHS = $\dfrac{{{{\sin }^2}{\rm{A}} + {\rm{\;}}{{\cos }^2}{\rm{A}}}}{{\sin {\rm{A}}\cos {\rm{A}}}}$
From trigonometric identity, ${\sin ^2}{\rm{A}} + {\rm{\;}}{\cos ^2}{\rm{A}} = 1$
LHS = $\dfrac{1}{{\sin {\rm{A}}\cos {\rm{A}}}} = {\rm{cosec\;A}}\sec {\rm{A}}$
As we know that $\sec \left( {90^\circ - {\rm{A}}} \right) = {\rm{cosecA}}$ and ${\rm{cosec}}\left( {90^\circ - {\rm{A}}} \right) = \sec {\rm{A}}$
LHS = $\sec \left( {90^\circ - {\rm{A}}} \right){\rm{cosec}}\left( {90^\circ - {\rm{A}}} \right)$ = RHS

Note: Whenever we face such types of questions we need to use the standard identities of trigonometry in order to get the LHS part equal to the RHS part.