Question

please do this two questions fast...please

Answer 1

1.

R.H.S = 1 + sec A

L.H.S = (sin A tan A ) / ( 1 - cos A )

Using identity,

⇒tan A = sin A / cos A

= [ sin A ( sin A / cos A ) ] / ( 1 - cos A )

= ( sin²A / cos A ) / ( 1 - cos A )

Using identity,

⇒ sin²A = 1 - cos²A

= [ ( 1 - cos²A ) / cos A ] / ( 1 - cos A )

= [ ( 1² - cos²A ) / cos A ] / ( 1 - cos A )

Using identity,

⇒( a² - b² ) = ( a + b ) ( a - b )

= [ ( 1 + cos A ) ( 1 - cos A ) / cos A ] / ( 1 - cos A )

= ( 1 + cos A ) / cos A

= ( 1 / cos A ) + ( cos A / cos A )

Using identity,

⇒ sec A = ( 1 / cos A )

= sec A + 1

= 1 + sec A ( R.H.S )

Proved !!

2.

R.H.S = ( 1 - tan A )² / ( 1 - cot A )²

Using identity,

⇒ ( a - b )² = ( a² + b² - 2ab )

= ( 1² + tan²A - 2 tan A ) / ( 1² + cot²A - 2 cot A )

= ( 1 + tan²A - 2 tan A ) / ( 1 + cot²A - 2 cot A )

Using identity,

⇒ ( 1 + tan²A ) = sec²A

and,

⇒ ( 1 + cot²A ) = cosec²A

= ( sec²A - 2 tan A ) / ( cosec²A - 2 cot A )

Using identity,

⇒ sec²A = ( 1 / cos²A )

⇒ tan A = ( sin A / cos A )

⇒cosec²A = ( 1 / sin² A )

⇒ cot A = cos A / sin A

= [ ( 1 / cos²A ) - 2( sin A / cos A ) ] / [ ( 1 / sin²A ) - 2 ( cos A / sin A ) ]

= [ ( 1 - 2 sin A cos A ) / cos²A ] / [ ( 1 - 2 sin A cos A ) / sin²A ]

= ( 1 / cos²A ) / ( 1 / sin²A )

= ( sin²A / cos²A )

= tan²A ( R.H.S )

L.H.S = ( 1 + tan²A ) / ( 1 + cot²A )

Using identity,

⇒ ( 1 + tan²A ) = sec²A

⇒ ( 1 + cot²A ) = cosec²A

= ( sec²A ) / ( cosec²A )

Using identity,

⇒ sec A = ( 1 / cos A )

⇒ cosec A = ( 1 / sin A )

= ( 1 / cos²A ) / ( 1 / sin²A )

= ( sin²A / cos²A )

= tan²A ( R.H.S )

Proved !!

Answer 2

◢GIVEN

$\frac{1 + {tan}^{2}a}{1 + {cot}^{2}a } = \frac{ {(1 - tana)}^{2} }{( {1 - cota)}^{2} }$
______________________________

◢TAKING LHS :-

$= > \frac{1 + {tan}^{2}a }{1 + {cot}^{2}a} = \frac{ {sec}^{2}a }{ {cosec}^{2}a} \\ \\ = \frac{ {sin}^{2} a}{ {cos}^{2}a } \\ \\ = {tan}^{2} a \: \: \: \: \: ........[Eq _{1}]$
_______________________________

◢NOW

┌◢TAKING RHS :-

$= > \frac{ {(1 - tana)}^{2} }{ {(1 - cota)}^{2} } = \frac{1 + {tan}^{2} a - 2tana}{1 + {cot}^{2} a - 2cota} \\ \\ = \frac{ {sec}^{2} a - 2tana}{ {cosec}^{2}a - 2cota } = \frac{ \frac{1}{ {cos}^{2}a } - \frac{2sina}{cosa} }{ \frac{1}{ {sin}^{2} a} - \frac{2cosa}{sina} } \\ \\ = \frac{ \frac{1}{cosa} ( \frac{1}{cosa} - 2sina)}{ \frac{1}{sina}( \frac{1}{sina} - 2cosa) } = \frac{sina( \frac{1 - 2sina.cosa}{cosa} )}{cosa( \frac{1 - 2sina.cosa}{sina} )} \\ \\ = \frac{ {sin}^{2} a}{ {cos}^{2}a } = \: {tan}^{2} a \: \: \: \: \: \: ..[Eq _{2}]$
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◢SO

$= > Eq _{1} = Eq _{2}$

◢HENCE

=> LHS = RHS ___________[◢PRPVED]

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