KnowledgeBoat Logo
|

Mathematics

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

(sec A - tan A)2(1 + sin A) = 1 - sin A.

Trigonometric Identities

21 Likes

Answer

Solving L.H.S.,

⇒(1cos A−sin Acos A)2(1+sin A)=(1 - sin Acos A)2(1 + sin A)=(1 - sin A)2(1 + sin A)1 - sin2A=(1 - sin A)2(1 + sin A)(1 - sin A)(1 + sin A)=1 - sin A.\Rightarrow \Big(\dfrac{1}{\text{cos A}} - \dfrac{\text{sin A}}{\text{cos A}}\Big)^2(1 + \text{sin A}) \\[1em] = \Big(\dfrac{\text{1 - sin A}}{\text{cos A}}\Big)^2\text{(1 + sin A)} \\[1em] = \dfrac{\text{(1 - sin A)}^2\text{(1 + sin A)}}{\text{1 - sin}^2 A} \\[1em] = \dfrac{\text{(1 - sin A)}^2\text{(1 + sin A)}}{\text{(1 - sin A)(1 + sin A)}} \\[1em] = \text{1 - sin A}.

Since, L.H.S. = R.H.S. hence proved that (sec A - tan A)2(1 + sin A) = 1 - sin A.

Answered By

12 Likes

Related Questions