Question

Please solve Expression (tanA/1-cotA) + (CotA/1-tanA) can be written as

1)sinA cosA +1
2)secA cosecA + 1
3)tanA + cotA
4)secA + cosecA ​

Answer 1

Answer:

2)secA cosecA + 1

Step-by-step explanation:

Given expression :

$\implies{\mathsf{ \frac{ \tan(a) }{1 - \cot(a) } + \frac{ \cot(a) }{1 - \tan(a) }}} \\ \\ \implies{\mathsf{ \frac{ \sin(a) }{ \cos(a) } \times \frac{1}{1 - \frac{ \cos(a) }{ \sin(a) } } + \frac{ \cos(a) }{ \sin(a) } \times \frac{1}{1 - \frac{ \sin(a) }{ \cos(a) } } }} \\ \\ \implies{\mathsf{ \frac{ \sin(a) }{ \cos(a) } \times \frac{1}{ \frac{ \sin(a) - \cos(a) }{ \sin(a) } } + \frac{ \cos(a) }{ \sin(a) } \times ( \frac{1}{ \frac{ \cos(a) - \sin(a) }{ \cos(a) } }) }} \\ \\ \implies{\mathsf{ \frac{ \sin(a) }{ \cos(a) } \times ( \frac{ \sin(a) }{ \sin(a) - \cos(a) } ) + \frac{ \cos(a) }{ \sin(a) } \times \frac{ \cos(a) }{ \cos(a) - \sin(a) }}}$

Taking negative(minus) common from 2nd term

$\implies{\mathsf{ \frac{ { \sin}^{2}(a)}{ \cos(a) \times ( \sin(a) - \cos(a) )} - \frac{ { \cos }^{2}(a) }{ \sin(a) \times ( \sin(a) - \cos(a)) }}}$

Taking L.C.M

$\implies{\mathsf{ \frac{1}{ \sin(a) - \cos(a) } \times ( \frac{ { \sin }^{3}(a) - { \cos }^{3}(a) }{ \cos(a) \sin(a) } )}}$

Using identity ,

$\implies$ A³ - B³ = (A - B)×(A² + AB + B²)

$\implies{\mathsf{ \frac{ \sin(a) - \cos(a) \times ( { \sin }^{2}(a) + \sin(a) \cos(a) + { \cos }^{2}(a)) }{( \sin(a) - \cos(a) ) \times ( \sin(a) \cos(a)) } }}$

By Canceling the terms

$\implies{\mathsf{ \frac{ { \sin}^{2}(a) + \sin(a) \cos(a) + { \cos}^{2} (a) }{ \cos(a) \sin(a) } }} \\ \\ \implies{\mathsf{ \frac{1 + \sin(a) \cos(a) }{ \sin(a) \cos(a) }}} \\ \\ \implies{\mathsf{ \frac{1}{ \sin(a) \cos(a) } + \frac{ \sin(a) \cos(a) }{ \sin(a) \cos(a) }}} \\ \\ \implies{\mathsf{ 1 + \sec(a) \csc(a)}}$

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$\underline{\bold{\mathtt{Identities\: Used}}}$:—

◾ A³ - B³ = (A - B)×(A² + AB + B²)

◾Sin²A + Cos²A = 1

◾TanA = SinA ÷ CosA