Question

tan/1-cotA +cotA/1-tanA =secAcosecA+1

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Answer 1

Answer:

if u give the options it's easy way to get the answer.

Answer 2

$\huge{\underline{\underline{\red{♡Solution→}}}}$

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

$\bold{\huge{\underline{\underline{\rm{ LHS :}}}}}$

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

$= \frac{ \frac{sin \: a}{cos \: a} }{1 - \frac{cos \: a}{sin \: a} } + \frac{ \frac{cos \: a}{sin \: a} }{1 - \frac{sin \: a}{cos \: a} } \\ = \frac{ \frac{sin \: a}{cos \: a} }{ \frac{sin \: a - cos \: a}{sin \: a} } + \frac{ \frac{cos \: a}{sin \: a} }{ \frac{cos \: a - sin \: a}{cos \: a} } \\ = \frac{ {sin \:}^{2} a}{cos \: a(sin \: a \: - cos \: a)} + \frac{ {cos}^{2}a }{sin \: a(cos \: a - sin \: a)}$

$= \frac{ {sin \:}^{2} a}{cos \: a(sin \: a \: - cos \: a)} - \frac{ {cos}^{2}a }{sin \: a(sin \: a - cos \: a)}$

$= \frac{ {sin}^{3}a - {cos}^{3} a}{sin \: a \: cos \: a(sin \: a - cos \: a)}$

$= \frac{(sin \: a - cos \: a)( {sin}^{2}a + {cos}^{2} a + sin \: a \: cos \: a) }{sin \: a \: cos \: a(sin \: a \: - cos \: a)}$

(Sin a - cos a) will be cancelled.

And sin²a + cos²a = 1

$= \frac{1 + sin \: a \: cos \: a}{sin \: a \: cos \: a}$

$= \frac{1}{sin \: a \: cos \: a} + \frac{sin \: a \: cos \: a}{sin \: a \: cos \: a}$

We know that →

• 1/sin a = cosec a
• 1/cos a = sec a

$= sec \: a \: + cosec \: a \: + 1$

$\bold{\huge{\underline{\underline{\rm{ RHS ↑}}}}}$

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