Prove the following identity:
tan2 A + cot2 A + 2 = sec2 A cosec2 A
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Solving L.H.S of equation,
⇒sin2Acos2A+cos2Asin2A+2⇒sin4A+cos4A+2cos2Asin2Acos2Asin2A⇒(sin2A+cos2A)2cos2Asin2A⇒12cos2Asin2A⇒1sin2A×1cos2A⇒cosec2Asec2A.\Rightarrow \dfrac{\sin^2 A}{\cos^2 A} + \dfrac{\cos^2 A}{\sin^2 A} + 2 \\[1em] \Rightarrow \dfrac{\sin^4 A + \cos^4 A + 2\cos^2 A \sin^2 A}{\cos^2 A \sin^2 A} \\[1em] \Rightarrow \dfrac{(\sin^2 A + \cos^2 A)^2}{\cos^2 A \sin^2 A} \\[1em] \Rightarrow \dfrac{1^2}{\cos^2 A \sin^2 A} \\[1em] \Rightarrow \dfrac{1}{\sin^2 A} \times \dfrac{1}{\cos^2 A} \\[1em] \Rightarrow \cosec^2 A \sec^2 A .⇒cos2Asin2A+sin2Acos2A+2⇒cos2Asin2Asin4A+cos4A+2cos2Asin2A⇒cos2Asin2A(sin2A+cos2A)2⇒cos2Asin2A12⇒sin2A1×cos2A1⇒cosec2Asec2A.
Since, L.H.S. = R.H.S.,
Hence, proved that tan2 A + cot2 A + 2 = sec2 A cosec2 A.
Answered By
cot2 A - cos2 A = cos2 A cot2 A
sec2 A + cosec2 A = sec2 A cosec2 A
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
