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`1/(tan3A+tanA)-1/(cot3A+cotA)=cot4A`
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`1/(tan3A+tanA)-1/(cot3A+cotA)=cot4A`
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How to solve 1/(tan3A+tanA) - 1/(cot3A+cotA) = cot4A trigonometry example PART-10
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Prove that: 1/(tan3A + tanA) -1/(cot3A+cotA) = cot4A
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Prove that: 1/(Tan3A + TanA) - 1/(Cot3A + CotA) = Cot4A
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Prove that: 1/(tan3A + tanA) - 1/(cot3A + cotA) = cot4A
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Prove that: 1/(tan3A + tanA) - 1/(cot3A + cotA) = cot4A
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Prove that 1/(tan3A-tana)-1/(cot3A-cotA) = cot2A
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Prove 1/(Tan3A+TanA) -1/(Cot3A+CotA) = Cot4A Class Ten || Trigonometry || Hindi ||
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Show that : 1/(tan 3A-tanA) - 1/(cot 3A - cot A) = cot 2A
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Prove that: CotA/(CotA-Cot3A)- TanA/Tan3A-TanA=1|| Prove that:1/Tan3A+TanA) -1/CotA+CotA=Cot4A || Rk
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Prove that: cotA/(cotA - cot3A) - tanA/(tan3A - tanA) = 1
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\( \frac{1}{\tan 3 A-\tan A}-\frac{1}{\cot 3 A-\cot A}= \) (A) \( \tan A \) (B) \( \tan 2 A \) (...
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Show that : `1/(tan 3 A + tan A) - 1/(cot 3A + cot A) = cot 4A`
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If in a `Delta ABC, tanA+tanB+tanC=0` then `cotA cotB cot C=` `a.6, b. 1, c. 1/6, d.` none of