Question

Prove that: (tan^3A/1+tan^2A) +(Cot^3A/1+Cot^2A) = SecA.CosecA - 2Sina.CosA​

Answer 1

Answer:

Step-by-step explanation:

Tan^3A / Sec^2A + Cot^3A / Cosec^2A

= (sin^3A/cos^3A) / (1 / Cos^2A) + (Cos^3A/Sin^3A) / (1 / Sin^2A)

= Sin^3A/CosA + Cos^3A/SinA

= (Sin^4A + Cos^4A) / SinA.CosA

= [ (Sin^2A + Cos^2A)^2 - 2Sin^2A.Cos^2A] / SinA.CosA

= ( 1- 2Sin^A.Cos^A)/ SinA.CosA

RHS = SecA CosecA - 2sinAcosA

= 1/CosA . 1/SinA - 2SinACosA

= (1 - Sin^2A.Cos^2A) / sinAcosA

Hence LHS = RHS (PROVED)

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Answer 2

       

$\boxed{ LHS } \\ \\ \boxed{\frac{\tan^3 A}{1+\tan^2 A}+\frac{\cot^3 A}{1+\cot^2 A}} \\ \\ \\ \boxed{\frac{(\frac{\sin A}{\cos A})^3}{\sec^2 A}+\frac{(\frac{\cos A}{\sin A})^3}{\csc^2 A}} \\ \\ \\ \boxed{\frac{\frac{\sin^3A}{\cos^3A}}{\sec^2A}+\frac{\frac{\cos^3A}{\sin^3A}}{\csc^2A}} \\ \\ \\ \boxed{(\frac{\sin^3A}{\cos^3A} \div \sec^2A)+(\frac{\cos^3A}{\sin^3A} \div \csc^2A)} \\ \\ \\ \boxed{(\frac{\sin^3A}{\cos^3A} \cdot \frac{1}{\sec^2A})+(\frac{\cos^3A}{\sin^3A} \cdot \frac{1}{\csc^2A}}$

$\boxed{(\frac{\sin^3A}{\cos^3A} \cdot \cos^2A)+(\frac{\cos^3A}{\sin^3A} \cdot \sin^2A)} \\ \\ \\ \boxed{\frac{\sin^3A}{\cos A}+\frac{\cos^3A}{\sin A}} \\ \\ \\ \boxed{\frac{\sin^4A+\cos^4A}{\cos A \cdot \sin A}} \\ \\ \\ \boxed{\frac{\sin^4A+\cos^4A+2\sin^2A \cdot \cos^2A-2\sin^2A \cdot \cos^2A}{\cos A \cdot \sin A}} \\ \\ \\ \boxed{\frac{(\sin^2 A+\cos^2 A)^2-2\sin^2 A \cdot \cos^2 A}{\cos A \cdot \sin A}} \\ \\ \\ \boxed{\frac{1^2-2\sin^2A \cdot \cos^2A}{\cos A \cdot \sin A}}$

$\boxed{\frac{1-2\sin^2A \cdot \cos^2A}{\cos A \cdot \sin A}} \\ \\ \\ \boxed{\frac{1}{\cos A \cdot \sin A}-\frac{2\sin^2A \cdot \cos^2 A}{\cos A \cdot \sin A}} \\ \\ \\ \boxed{\frac{1}{\cos A} \cdot \frac{1}{\sin A}-\frac{2(\sin A \cdot \cos A)^2}{\cos A \cdot \sin A}} \\ \\ \\ \boxed{\sec A \cdot \csc A - 2\sin A \cdot \cos A} \\ \\ \boxed{ RHS }$

$Hence proved!$

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