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Question 5 (iv)
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(iv) (1+secA)secA=sin2A(1cosA)
[Hint : Simplify L.H.S and R.H.S separately]


Solution

(iv) (1+secA)secA=sin2A(1cosA)L.H.S. =(1+secA)secA=(1+1cosA)1cosA=(cosA+1)cosA1cosA=cosA+1R.H.S. =sin2A(1cosA)=(1cos2A)(1cosA)=(1cosA)(1+cosA)(1cosA)=cosA+1L.H.S.=R.H.S.

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