Question

solve this question very quickly

Answer 1

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Answer 2

Question:

${tan}^{2} A + {cot}^{2} A = {sec}^{2} A {cosec}^{2} A -2$

Solution:

To prove: ${tan}^{2} A + {cot}^{2} A = {sec}^{2} A {cosec}^{2} A -2$

Proof:

LHS = ${tan}^{2} A + {cot}^{2} A$

RHS = ${sec}^{2} A {cosec}^{2} A -2$

LHS = ${tan}^{2} A + {cot}^{2} A$

$= {sec}^{2} A - 1 + {cosec}^{2} A - 1$

$= {sec}^{2} A + {cosec}^{2} A - 2$

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

$= \dfrac{{sin}^{2} A + {cos}^{2} A}{{cos}^{2} A {sin}^{2}A} - 2$

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

$= {sec}^{2} A {cosec}^{2} A - 2$ = RHS

Identities used

• $sec\theta = \dfrac{1}{cos\theta}$

• $cosec\theta = \dfrac{1}{sin\theta}$

• ${sin}^{2} \theta + {cos}^{2} \theta = 1$