Question

$sec ^{2}( \alpha ) - \tan( \alpha ) \div \sec ^{2} + \tan( \alpha )$
​

Answer 1

Answer:

$\dfrac{ \sec^{2} ( \alpha ) - \tan( \alpha ) }{ \sec^{2} ( \alpha ) + \tan( \alpha ) }$

We know that -

• secθ = 1/cosθ
• tanθ = sinθ/cosθ

On substituting these value in above -

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

On taking LCM,

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

On changing division into multiplication.

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

On cancelling cos (α),

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