Question

Answer the question

Answer 1

step by step explanation:

trigonometry;

we have to prove :

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

LHS

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

we know that

$\sec {}^{2} ( \alpha ) - \tan {}^{2} ( \alpha ) = 1$

then ,

LHS

$= \frac{( \sec( \alpha ) - \tan( \alpha ) )( \sec {}^{2} ( \alpha ) - \tan {}^{2} ( \alpha ) )}{1 + \sec( \alpha ) - \tan( \alpha ) }$

[ use( a+b) ( a-b) = a^2 - b^2]

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

$= \sec( \alpha ) - \tan( \alpha )$

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

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

$= rhs$

I hope it helps you

Answer 2

ANSWER:-

➣

• To Prove :

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

SOLUTION:-

• LHS

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

As we know:-

$\sec {}^{2} ( \alpha ) - \tan {}^{2} ( \alpha )= 1$

Therefore ,

LHS

$= \frac{( \sec( \alpha ) - \tan( \alpha ) )( \sec {}^{2} ( \alpha ) - \tan {}^{2} ( \alpha ) )}{1 + \sec( \alpha ) - \tan( \alpha ) }$

We know:-

$\boxed{\bf{( a+b) ( a-b) = a^2 - b^2}}$

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

$= \sec( \alpha ) - \tan( \alpha )$

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

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

• RHS:-

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

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L.H.S = R.H.S

Hence Proved ✅