Question

to prove : COT A/1-TAN A = COT A-1/2-SEC² A ​

Answer 1

Answer:

Formula used (tan^2)x=(sec^2)x + 1

Step-by-step explanation:

Please refer to the attachment for explanation....

Answer 2

■■■■■■■■CORRECT QUESTION■■■■■■■

$\frac{cot \: a}{1 - tan \: a \: } = \frac{cot \: a \: + 1}{2 - {sec}^{2}a }$

✿✿✿✿✿✿✿FORMULA USED✿✿✿✿✿✿✿

$trigonometric \: identities \\ \\ 1 + {tan}^{2} \theta \: = {sec}^{2} \theta \: \\ \\ and \\ \\ cot \theta \: = \frac{1}{tan \theta} \\ \\ with \\ \\ algebraic \: identity \: ( {p}^{2} - {q}^{2} ) = (p + q)(p - q)$

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LHS=

$\frac{cot \: a}{1 - tan \: a } \\ \\ = \frac{1}{tan \: a} \times \: \frac{1}{1 -tan \: a} \\ \\ = \frac{1}{tan \: a \: - {tan}^{2} a}$

RHS=

$\frac{ cot \: a \: + 1}{2 - {sec}^{2}a } \\ \\ = ( \frac{1}{tan \: a} \: + \frac{1}{1} ) \div (2 - {sec}^{2} a) \\ \\ = ( \frac{1 + tan \: a}{tan \: a} ) \div (2 - (1 + {tan}^{2} a)) \\ \\ = (\frac{1 + tan \: a}{tan \: a} ) \div (2 - 1 - {tan}^{2} a) \\ \\ = ( \frac{1 + tan \: a}{tan \: a} ) \div (1 - {tan}^{2} a) \\ \\ = ( \frac{1 + tan \: a}{tan \: a} ) \times (\frac{1}{1 - {tan}^{2}a } ) \\ \\ = ( \frac{1 + tan \: a}{tan \: a} ) \times ( \frac{1}{(1 + tan \: a)(1 - tan \: a)} ) \\ \\ = \frac{1}{(tan \: a)(1 - tan \: a)} \\ \\ = \frac{1}{tan \: a - {tan}^{2}a }$

LHS = RHS, HENCE PROVED

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It is another way to solve this question other than the way given by above user

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