Question

Prove that: cosec^3 A - cot^3 A = 1 + 3cosec^2 A.cot^2 A

Answer 1

Answer:

question is wrong type,your question will be cosec^6A-cot^6A=1+3cosec^2A.cot^2A then you will be solve.

Thanks!

Answer 2

Here the LHS should be equal to RHS

The correct question is $Cosec^{6}A - Cot^{6}A = 1 + 3Cosec^{2}A.Cot^{2}A$

Given:

$Cosec^{6}A - Cot^{6}A = 1 + 3Cosec^{2}A.Cot^{2}A$

To find:

To prove that LHS = RHS.

Solution:

Let us look at the equation that is $Cosec^{6}A - Cot^{6}A = 1 + 3Cosec^{2}A.Cot^{2}A$

Let us transpose this equation in such a way that LHS has the trigonometric functions and RHS has the value 1. This can be done by changing the position of the function and its sign.

$Cosec^{6}A - Cot^{6}A = 1 + 3Cosec^{2}A.Cot^{2}A$$Cosec^{6}A - Cot^{6}A - 3Cosec^{2}A.Cot^{2}A = 1$

The power of a value can be written as its multiple. Sine 6 is a power here it can be written as 2x3.

$(Cosec^{2})^{3}A - (Cot^{2})^3A - 3Cosec^{2}A.Cot^{2}A = 1$

As we know that A³ - B³ = (A - B)(A² + AB + B²)

So here

$(Cosec^{2})^{3}A - (Cot^{2})^3A = (1) ((Cosec^2A)^2 + Cosec^2ACot^2A + (Cot^2A)^2)$

Through trigonometric formulas, we know that $Cosec^{2}A - Cot^{2}A = 1$

Substitute this in that equation,

$(Cosec^{2})^{3}A - (Cot^{2})^3A = (1) ((Cosec^2A)^2 + Cosec^2ACot^2A + (Cot^2A)^2)$

Add this to the LHS of the proving equation,

$(Cosec^{2})^{3}A - (Cot^{2})^3A - 3Cosec^{2}A.Cot^{2}A$

$(Cosec^2A)^2 + Cosec^2ACot^2A + (Cot^2A)^2 - 3Cosec^2A.Cot^2A$

Here we are going to add $2Cosec^2A.Cot^2A -2Cosec^2A.Cot^2A$

$(Cosec^2A)^2 + Cosec^2ACot^2A + (Cot^2A)^2 + 2Cosec^2A.Cot^2A -2Cosec^2A.Cot^2A - 3Cosec^2A.Cot^2A$

$(Cosec^2A)^2 + Cosec^2ACot^2A + 2Cosec^2A.Cot^2A + (Cot^2A)^2 -2Cosec^2A.Cot^2A - 3Cosec^2A.Cot^2A$$(Cosec^2A)^2 + (Cot^2A)^2 -2Cosec^2A.Cot^2A + 3Cosec^2A.Cot^2An - 3Cosec^2A.Cot^2A$As we know (A² - B²) = A² + B² - 2AB

$(Cosec^2A)^2 + (Cot^2A)^2 -2Cosec^2A.Cot^2A$ can be written as,$(Cosec^{2}A)^2 - (Cot^{2}A)^2$

According to trigonometric formulas, we know that $Cosec^{2}A - Cot^{2}A = 1$

= 1

Therefore,  LHS = RHS.

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