KnowledgeBoat Logo
|

Mathematics

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:

1 + sin A1 - sin A=tan A + sec A\sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}} = \text{tan A + sec A}

Trigonometric Identities

43 Likes

Answer

The L.H.S. of the equation can be written as,

⇒(1 + sin A)(1 + sin A)(1 - sin A)(1 + sin A)⇒(1 + sin A)2(1 - sin2A)⇒(1 + sin A)2cos2A⇒1 + sin Acos A⇒1cos A+sin Acos A⇒sec A + tan A\Rightarrow \sqrt{\dfrac{\text{(1 + sin A)(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}}} \\[1em] \Rightarrow \sqrt{\dfrac{\text{(1 + sin A)}^2}{\text{(1 - sin}^2 A)}} \\[1em] \Rightarrow \sqrt{\dfrac{\text{(1 + sin A)}^2}{\text{cos}^2 A}} \\[1em] \Rightarrow \dfrac{\text{1 + sin A}}{\text{cos A}} \\[1em] \Rightarrow \dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}} \\[1em] \Rightarrow \text{sec A + tan A}

Since, L.H.S. = R.H.S. hence, proved that 1 + sin A1 - sin A\sqrt{\dfrac{\text{1 + sin A}}{\text{1 - sin A}}} = tan A + sec A

Answered By

17 Likes

Related Questions