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Question

If cotθ=3x112x then show that cscθ+cotθ=6x or 16x.

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Solution

cotθ=3x112x
Squaring both sides, we have
cot2θ=(3x112x)2
csc2θ1=9x2+1144x212
csc2θ=9x2+1144x212+1
csc2θ=9x2+1144x2+12
csc2θ=(3x+112x)2
cscθ=±(3x+112x)
Now,
Case I:- cscθ=3x+112x
Therefore,
cotθ+cscθ=(3x112x)+(3x+112x)=6x
Case II:- cscθ=(3x+112x)
Therefore,
cotθ+cscθ=(3x112x)+(3x112x)=16x
Hence proved.

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