Question

Cosec6A-cot6A=1+3cot2Acosec2A

Answer 1

Answer:

\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.  

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