KnowledgeBoat Logo

Mathematics

Prove that :

cos2A+tan2A−1sin2A=tan2A\dfrac{\text{cos}^2 A + \text{tan}^2 A - 1}{\text{sin}^2 A} = \text{tan}^2 A.

Trigonometric Identities

1 Like

Answer

To prove:

cos2A+tan2A−1sin2A=tan2A\dfrac{\text{cos}^2 A + \text{tan}^2 A - 1}{\text{sin}^2 A} = \text{tan}^2 A

By formula,

cos2 A = 1 - sin2 A

Solving L.H.S. of the above equation :

⇒cos2A+tan2A−1sin2A⇒1 - sin2A+sin2Acos2A−1sin2A⇒1−sin2A+sin2Acos2A−1sin2A⇒-sin2A+sin2A sec2Asin2A⇒sin2A(−1+sec2A)sin2A⇒sec2A−1⇒tan2A.\Rightarrow \dfrac{\text{cos}^2 A + \text{tan}^2 A - 1}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{1 - sin}^2 A + \dfrac{\text{sin}^2 A}{\text{cos}^2 A} - 1}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\cancel{1} - \text{sin}^2 A + \dfrac{\text{sin}^2 A}{\text{cos}^2 A} - \cancel{1}}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{-sin}^2 A + \text{sin}^2 A\text{ sec}^2 A}{\text{sin}^2 A} \\[1em] \Rightarrow \dfrac{\text{sin}^2 A(-1 + \text{sec}^2 A)}{\text{sin}^2 A} \\[1em] \Rightarrow \text{sec}^2 A - 1 \\[1em] \Rightarrow \text{tan}^2 A.

Since, L.H.S. = R.H.S.

Hence, proved that cos2A+tan2A−1sin2A=tan2A\dfrac{\text{cos}^2 A + \text{tan}^2 A - 1}{\text{sin}^2 A} = \text{tan}^2 A.

Answered By

1 Like

Related Questions