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Question: \(\dfrac{{\sin {\rm{A}}}}{{\sin \left( {90^\circ - {\rm{A}}} \right)}} + {\rm{\;}}\dfrac{{\cos {\rm{...

sin⁡Asin⁡(90∘−A)+  cos⁡Acos⁡(90∘−A)=  sec⁡(90∘−A)cosec(90∘−A)\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)

Explanation

Solution

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⁡(90∘−A)=cos⁡A\sin \left( {90^\circ - {\rm{A}}} \right) = \cos {\rm{A}} and cos⁡(90∘−A)=sin⁡A\cos \left( {90^\circ - {\rm{A}}} \right) = \sin {\rm{A}} and then make use of identity sin⁡2A+  cos⁡2A=1{\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 = sin⁡Asin⁡(90∘−A)+  cos⁡Acos⁡(90∘−A)\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⁡(90∘−A)=cos⁡A\sin \left( {90^\circ - {\rm{A}}} \right) = \cos {\rm{A}} and cos⁡(90∘−A)=sin⁡A\cos \left( {90^\circ - {\rm{A}}} \right) = \sin {\rm{A}}
LHS = sin⁡Acos⁡A+  cos⁡Asin⁡A\dfrac{{\sin {\rm{A}}}}{{\cos {\rm{A}}}} + {\rm{\;}}\dfrac{{\cos {\rm{A}}}}{{\sin {\rm{A}}}}
Taking LCM, we get
LHS = sin⁡2A+  cos⁡2Asin⁡Acos⁡A\dfrac{{{{\sin }^2}{\rm{A}} + {\rm{\;}}{{\cos }^2}{\rm{A}}}}{{\sin {\rm{A}}\cos {\rm{A}}}}
From trigonometric identity, sin⁡2A+  cos⁡2A=1{\sin ^2}{\rm{A}} + {\rm{\;}}{\cos ^2}{\rm{A}} = 1
LHS = 1sin⁡Acos⁡A=cosec  Asec⁡A\dfrac{1}{{\sin {\rm{A}}\cos {\rm{A}}}} = {\rm{cosec\;A}}\sec {\rm{A}}
As we know that sec⁡(90∘−A)=cosecA\sec \left( {90^\circ - {\rm{A}}} \right) = {\rm{cosecA}} and cosec(90∘−A)=sec⁡A{\rm{cosec}}\left( {90^\circ - {\rm{A}}} \right) = \sec {\rm{A}}
LHS = sec⁡(90∘−A)cosec(90∘−A)\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.