Question

one upon sec theta minus 10 theta minus one upon cos theta is equals to one upon cos theta minus one upon sec theta + tan theta​

Answer 1

AnswEr :

• To Prove :

$\small \dfrac{1}{ \sec( \alpha ) - \tan( \alpha ) } - \dfrac{1}{ \cos( \alpha ) } = \dfrac{1}{ \cos( \alpha ) } - \dfrac{1}{ \sec( \alpha ) + \tan( \alpha ) }$

• Proof :

I'm taking $(\alpha)$ in the place of $(\theta)$

◑ Taking LHS First ;

$\longrightarrow\dfrac{1}{ \sec( \alpha ) - \tan( \alpha ) } - \dfrac{1}{ \cos( \alpha ) }$

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

$\longrightarrow\dfrac{ \cos( \alpha ) }{ 1 - \sin( \alpha ) } - \dfrac{1}{ \cos( \alpha ) }$

$\longrightarrow\dfrac{ \cos^{2} ( \alpha ) - 1}{ (1 - \sin( \alpha ))\cos( \alpha ) }$

$\longrightarrow\dfrac{ \cos^{2} ( \alpha ) - 1 + \sin( \alpha ) }{ (1 - \sin( \alpha )) \cos( \alpha )}$

$\longrightarrow\dfrac{ - ( - \cos^{2} ( \alpha ) + 1 - \sin( \alpha )) }{ (1 - \sin( \alpha )) \cos( \alpha )}$

$\longrightarrow\dfrac{ - ( (1- \cos^{2} ( \alpha ) ) - \sin( \alpha )) }{ (1 - \sin( \alpha ))\cos( \alpha ) }$

$\longrightarrow\dfrac{ - ( \sin^{2} ( \alpha ) - \sin( \alpha )) }{ (1 - \sin( \alpha ))\cos(\alpha )}$

$\longrightarrow\dfrac{ \sin( \alpha ) - \sin^{2} ( \alpha ) }{ (1 - \sin( \alpha ))\cos( \alpha )}$

$\longrightarrow\dfrac{ \cancel{(1 - \sin( \alpha ))}( \sin( \alpha )) }{ \cancel{(1 - \sin( \alpha ))}( \cos( \alpha )) }$

$\longrightarrow\dfrac{ \sin( \alpha ) }{ \cos( \alpha ) }$

$\longrightarrow \large \tan( \alpha )$

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◑ Now Taking RHS ;

$\longrightarrow \dfrac{1}{ \cos( \alpha ) } - \dfrac{1}{ \sec( \alpha ) + \tan( \alpha ) }$

$\longrightarrow \dfrac{1}{ \cos( \alpha ) } - \dfrac{1}{ \frac{1}{ \cos( \alpha ) } + \frac{ \sin( \alpha ) }{ \cos( \alpha ) } }$

$\longrightarrow \dfrac{1}{ \cos( \alpha ) } - \dfrac{ \cos( \alpha ) }{ 1 + \sin( \alpha ) }$

$\longrightarrow \dfrac{1 + \sin( \alpha ) - \cos^{2} ( \alpha ) }{ \cos( \alpha )( 1 + \sin( \alpha ) ) }$

$\longrightarrow \dfrac{ \sin( \alpha ) + (1 - \cos^{2} ( \alpha ) ) }{ \cos( \alpha )( 1 + \sin( \alpha ) ) }$

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

$\longrightarrow \dfrac{ \sin( \alpha ) \cancel{( 1 + \sin ( \alpha ))} }{ \cos( \alpha )\cancel{( 1 + \sin( \alpha ) )} }$

$\longrightarrow\dfrac{ \sin( \alpha ) }{ \cos( \alpha )}$

$\longrightarrow \large \tan( \alpha )$

⠀

$\therefore\small\boxed{ \dfrac{1}{ \sec( \alpha ) - \tan( \alpha ) } - \dfrac{1}{ \cos( \alpha ) } = \dfrac{1}{ \cos( \alpha ) } - \dfrac{1}{ \sec( \alpha ) + \tan( \alpha ) } }$

Answer 2

I THINK THIS METHOD IS EASIER .