Question

(cosec^6A-cot^6A)=3cot^2Acosec^2A+1
Prove

Answer 1

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Answer 2

$\bold{\left(\left(\csc ^{2} A\right)^{3}-\left(\cot ^{2} A\right)^{3}\right)=3 \cot ^{2} A \cdot \csc ^{2} A+1}$

Solution:

First let us convert $\csc ^{2} A \text { into } 1+\cot ^{2}$ A after this place $1 + cot^2 A$ in place of $\csc^{2} of \left(\left(\csc ^{2} A\right)^{3}-\left(\cot ^{2} A\right)^{3}\right)$we get $\left(\left(1+\cot ^{2} A\right)^{3}-\left(\cot ^{2} A\right)^{3}\right)$

So, $\left(\left(1+\cot ^{2} A\right)^{3}-\left(\cot ^{2} A\right)^{3}\right)=3 \cot ^{2} A \cdot \csc ^{2} A+1$

Place $\cot ^{2} A=a,{in}\left(\left(1+\cot ^{2} A\right)^{3}-\left(\cot ^{2} A\right)^{3}\right)$  we get  

$\begin{array}{l}{\left((1+a)^{3}-(a)^{3}\right)=\left(\left(1^{3}+a^{3}+3 a(1+a)\right)-(a)^{3}\right)} \\ \\{\left(\left(1^{3}+a^{3}+3 a(1+a)\right)-(a)^{3}\right)=1+3 a(a+1)}\end{array}$

Putting back $a=\cot ^{2} A$  

$\bold{1+3 a(a+1)=1+3 \cot ^{2} A\left(\cot ^{2} A+1\right)}$

$\bold{1+3 a(a+1)=1+3 \cot ^{2} A. \csc^{2} A}$

Hence LHS = RHS.