if tan2Θ=cot(Θ+6) where, 2Θ and Θ+6 are acute angles, find the value of Θ
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tan2Ф=cot(Ф+6)
we know that tan2Ф=cot(90-2Ф)
so cot(90-2Ф)=cot(Ф+6)
90-6=Ф+2Ф
84=3Ф
⇔Ф=28
we know that tan2Ф=cot(90-2Ф)
so cot(90-2Ф)=cot(Ф+6)
90-6=Ф+2Ф
84=3Ф
⇔Ф=28
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