Prove the following identity:
sec A (1 - sin A)(sec A + tan A) = 1
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Solving L.H.S. of the equation :
⇒ sec A(1 - sin A)(sec A + tan A)
⇒1cosA×(1−sinA)×(1cosA+sinAcosA)⇒1−sinAcosA×1+sinAcosA⇒1−sin2Acos2A⇒cos2Acos2A⇒1.\Rightarrow \dfrac{1}{\cos A} \times (1 - \sin A) \times \Big(\dfrac{1}{\cos A} + \dfrac{\sin A}{\cos A}\Big) \\[1em] \Rightarrow \dfrac{1 - \sin A}{\cos A} \times \dfrac{1 + \sin A}{\cos A} \\[1em] \Rightarrow \dfrac{1 - \sin^2 A}{\cos^2 A} \\[1em] \Rightarrow \dfrac{\cos^2 A}{\cos^2 A} \\[1em] \Rightarrow 1.⇒cosA1×(1−sinA)×(cosA1+cosAsinA)⇒cosA1−sinA×cosA1+sinA⇒cos2A1−sin2A⇒cos2Acos2A⇒1.
Since, L.H.S. = R.H.S.
Hence, proved that sec A (1 - sin A)(sec A + tan A) = 1.
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(sin2 θ - 1) (tan2 θ + 1) + 1 = 0
cosec A (1 + cos A)(cosec A - cot A) = 1
(cosec θ - sin θ)(sec θ - cos θ)(tan θ + cot θ) = 1
(cosec A + sin A)(cosec A - sin A) = cot2 A + cos2 A
