Prove the following identity:
(11+sinA)+(11−sinA)=2sec2A\Big(\dfrac{1}{1 + \sin A}\Big) + \Big(\dfrac{1}{1 - \sin A}\Big) = 2 \sec^2 A(1+sinA1)+(1−sinA1)=2sec2A
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Solving L.H.S. of the equation :
⇒11+sinA+11−sinA⇒1−sinA+1+sinA(1−sinA)(1+sinA)⇒2(1−sin2A) By formula, sin2A+cos2A=1⇒2cos2A⇒2sec2A.\Rightarrow \dfrac{1}{1 + \sin A} + \dfrac{1}{1 - \sin A} \\[1em] \Rightarrow \dfrac{1 - \sin A + 1 + \sin A}{(1 - \sin A)(1 + \sin A)} \\[1em] \Rightarrow \dfrac{2}{(1 - \sin^2 A)} \\[1em] \text{ By formula, } \sin^2 A + \cos^2 A = 1 \\[1em] \Rightarrow \dfrac{2}{\cos^2 A} \\[1em] \Rightarrow 2\sec^2 A.⇒1+sinA1+1−sinA1⇒(1−sinA)(1+sinA)1−sinA+1+sinA⇒(1−sin2A)2 By formula, sin2A+cos2A=1⇒cos2A2⇒2sec2A.
Since, L.H.S. = R.H.S.,
Hence, proved that (11+sinA)+(11−sinA)=2sec2A\Big(\dfrac{1}{1 + \sin A}\Big) + \Big(\dfrac{1}{1 - \sin A}\Big) = 2 \sec^2 A(1+sinA1)+(1−sinA1)=2sec2A.
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sin A (1 + tan A) + cos A (1 + cot A) = sec A + cosec A
(11+tan2A)+(11+cot2A)=1\Big(\dfrac{1}{1 + \tan^2 A}\Big) + \Big(\dfrac{1}{1 + \cot^2 A}\Big) = 1(1+tan2A1)+(1+cot2A1)=1
(1+sinAcosA)+(cosA1+sinA)=2secA\Big(\dfrac{1 + \sin A}{\cos A}\Big) + \Big(\dfrac{\cos A}{1 + \sin A}\Big) = 2 \sec A(cosA1+sinA)+(1+sinAcosA)=2secA
(cosecAcosecA−1)+(cosecAcosecA+1)=2sec2A\Big(\dfrac{\cosec A}{\cosec A - 1}\Big) + \Big(\dfrac{\cosec A}{\cosec A + 1}\Big) = 2 \sec^2 A(cosecA−1cosecA)+(cosecA+1cosecA)=2sec2A
