Question

tanA/1-cotA=cotA/1-tanA=1+secA.cosecA

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}=cosecAsecA+1$

Step-by-step explanation:

$LHS =\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}$$=\frac{\frac{1}{cotA}}{1-cotA}+\frac{cotA}{1-\frac{1}{cotA}}$

$=\frac{1}{cotA(1-cotA)}-\frac{cot^{2}A}{1-cotA}$

$=\frac{1-cot^{3}A}{cotA(1-cotA)}$

$=\frac{1^{3}-cot^{3}A}{cotA(1-cotA)}$

$=\frac{(1-cotA)(1^{2}+1\times cotA+cot^{2}A)}{cotA(1-cotA)}$

$=\frac{(1+cot^{2}A)+cotA}{cotA}$

$=\frac{cosec^{2}A+cotA}{cotA}$

$=\frac{cosec^{2}A}{cotA}+\frac{cotA}{cotA}$

$=\frac{\frac{1}{sin^{2}A}}{\frac{cosA}{sinA}}+1\\=\frac{1}{sinAcosA}+1\\=cosecAsecA+1\\=RHS$

Therefore,.

$\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}=cosecAsecA+1$

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