Question

show that (1 + cot^2 A) (1 - cos A) (1 + cos A) = 1​

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

$(1 + {cot}^{2} a)(1 - cos \: a)(1 + cos \: a) \\ \\ = (1 + cot ^{2} a)(1 - {cos}^{2} a) \\ \\ = 1 - {cos}^{2} a + {cot}^{2} a - {cot}^{2} a. {cos}^{2} a \\ \\ = 1 - {cos}^{2} a + \frac{ {cos}^{2} a}{ {sin}^{2} a} - \frac{ {cos}^{2} a}{sin ^{2}a } .{cos}^{2} a \\ \\ = \frac{ {sin}^{2} a - {cos}^{2} a. {sin}^{2} a + {cos}^{2}a - {cos}^{4} a } { {sin}^{2} a} \\ \\ = \frac{ {sin}^{2}a + {cos}^{2}a - {cos}^{2} a {sin}^{2} a - {cos}^{4} a }{ {sin}^{2} a} \\ \\ = \frac{1 - {cos}^{2}a( {sin}^{2}a + {cos}^{2} a) }{ {sin}^{2} a} \\ \\ = \frac{1 - {cos}^{2} a}{ {sin}^{2} a} \\ \\ = \frac{ {sin}^{2}a }{ {sin}^{2} a} \\ \\ = 1 (proved)\\ \\ \\ \\ formula - \: \: \: \: cot \: a \: = \frac{cos \: a}{sin \: a} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {sin}^{2} a + {cos}^{2} a = 1 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1 - {cos}^{2} a = {sin}^{2} a$