Question

Please answer step by step with attachment

Answer 1

AnSwEr

To prove :- 1/{sec∅+tan∅} = {1-sin∅}/Cos∅

Proof

Breaking sec∅ and tan∅ in sin or cos

LHS

=>1/{1/cos ∅ + sin∅ / cos ∅}

=>Cos∅/ Sin∅ + 1

Here multiplying with 1-sin∅

=>Cos∅ / Sin∅ + 1 × 1-sin∅ / 1-sin∅

=>cos∅(1-sin∅)/1-sin²∅

=>cos∅(1-sin∅)/cos²∅

=>1-sin∅ / cos∅

LHS = RHS (verified)

Answer 2

QUESTION:

$\frac{1}{ \sec( \alpha ) + \tan( \alpha ) } = \frac{1 - \sin( \alpha ) }{ \cos( \alpha ) }$

$\red {( \: I \: let \: thetha \: as \: \alpha )}$

ANSWER:

We use the trigonometry identity here;

1.

$\sec( \alpha ) = \frac{1}{ \cos( \alpha ) }$

2.

$\tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$

3.

${ \sin( \alpha ) }^{2} + { \cos( \alpha ) }^{2} = 1$

now come to main question;

TAKING LHS SIDE :

$\frac{1}{ \sec( \alpha ) + \tan( \alpha ) } \\ \frac{1}{ \frac{1}{ \cos( \alpha ) + \frac{ \sin( \alpha ) }{ \cos( \alpha ) } } } \\ \frac{1}{ \frac{1 + \sin( \alpha ) }{ \cos( \alpha ) } } \\ \frac{ \cos( \alpha ) }{1 + \sin( \alpha ) }$

$\red {[multiplying \: with \: 1 - \sin( \alpha ) ]}$

$\frac{ \cos( \alpha ) }{1 + \sin( \alpha ) } \times \frac{1 - \sin( \alpha ) }{1 - \sin( \alpha ) } \\ \frac{ \cos( \alpha ) \times (1 - \sin( \alpha ) }{ ({1 - \sin( \alpha ) })^{2} }$

$\frac{ \cos( \alpha ) (1 - \sin( \alpha ) }{ { \cos( \alpha ) }^{2} } \\ \red{(cos \: \alpha \: cancelled)}\: out) \\ \frac{1 - \sin( \alpha ) }{ \cos( \alpha ) }$

LHS = RHS

HENCE PROVED.

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