Question

Prove that tan(90-A)cotA/cosec2 A-cos2A=0.Explain.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$\frac{tan(90-A)cotA}{cosec^{2}A} - cos^{2}A=0$

Step-by-step explanation:

$LHS = \frac{tan(90-A)cotA}{cosec^{2}A} - cos^{2}A$

$=\frac{cotA\cdot cotA}{cosec^{2}A} - cos^{2}A$

/* tan(90-A) = cotA */

$= \frac{cot^{2}A}{cosec^{2}A}-cos^{2}A$

$=\frac{\frac{cos^{2}A}{sin^{2}A}}{\frac{1}{sin^{2}A}}-cos^{2}A$

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$i)cotA=\frac{cosA}{sinA}\\ii) cosecA = \frac{1}{sinA}$

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$=\cos^{2}A-cos^{2}A\\=0\\=RHS$

Therefore,

$\frac{tan(90-A)cotA}{cosec^{2}A} - cos^{2}A=0$

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