Question 5 (vi) Prove the following identities, where the angles involved are acute angles for which the expressions is defined. sec A + tan A = √(1+sinA1−sinA)
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Solution
√(1+sinA1−sinA)=R.H.SDividing R.H.S by cos A:=
⎷(1cosA+sinAcosA1cosA−sinAcosA)=√(secA+tanAsecA−tanA)=√(secA+tanAsecA−tanA)×(secA+tanAsecA+tanA)=secA+tanA√sec2A−tan2A=secA+tanA1(∵sec2A−tan2A=1)=secA+tanA=L.H.S⇒L.H.S=R.H.SProved.