Question

hello frnds please help me by solving this

Answer 1

Let angle A be 'x'
LHS=
$\frac{ \tan(x) }{ \sec(x) + 1} - \frac{ \tan(x) }{ - ( \sec(x) - 1) }$
ie
$\tan(x) ( \frac{1}{ \sec(x) + 1 } + \frac{1}{ \sec(x) - 1} )$
ie
$\tan(x) ( \frac{2 \sec(x) }{ { \sec(x) }^{2} - 1 } )$
ie
$\tan(x)( \frac{2 \sec(x) }{ { \tan(x) }^{2} } )$
Since sec^2 x-1=tan^2 x
ie
$\frac{2 \sec(x) }{ \tan(x) } = 2 \times \frac{1}{ \cos(x) } \times \frac{ \cos(x) }{ \sin(x) } = \frac{2}{ \sin(x) }$
ie
$2 \csc(x) = rhs$