Question

Trignometry pls solve it​

Answer 1

RHS

( sec^2A -1)/ (sec A -1 )^2

(sec A )^2 - (1)^2 / (Sec A -1)^2

(Sec A +1 ) × ( sec A -1 ) / (Sec A -1 ) × (sec A -1)

(sec A + 1 ) / ( sec A - 1)

( 1 + cos A ) / ( 1 - cos A )

Answer 2

Step-by-step explanation:

${\underline{{\text{To Prove:}}}}$

$\dfrac{1 + \cos }{1 - \cos} = \dfrac{ { \tan}^{2} }{( { \sec - 1)}^{2} }$

${\underline{{\text{Solution:}}}}$

L.H.S

$\dfrac{1 + \cos }{1 - \cos} \\ \\ = \frac{ 1 + \cos }{1 - \frac{1}{ \sec } } \\ \\ = \frac{1 + \cos}{ \frac{ \sec - 1 }{ \sec} } \\ \\ = \frac{(1 + \cos) \sec }{ \sec - 1} \\ \\ = \frac{ (\sec + \cos. \sec) }{( \sec - 1)} \: \: \: [ \text{using\: i}] \\ \\ = \frac{ (\sec + 1) ( \sec - 1)}{( \sec - 1)( \sec - 1)} \\ \\ \frac{ { \sec}^{2} - {1}^{2} }{ { (\sec - 1)}^{2} } \: \: [ \text{using(a+b)(a-b)} = {a}^{2} - {b}^{2} ] \\ \\ \frac{ \tan {}^{2} }{ { (\sec - 1)}^{2} } \: \: [ \text{using ii} ] \: \\ \\ \therefore \text{L.H.S = R.H.S [Proved]}$

${{\boxed{ \text{\blue{Some important trigonometry Rules}}}}}$

$sin \theta . cosec \theta = 1 \\ \\</p><p>\cos \theta. \sec\theta = 1.......[ \text{i}] \\ \\</p><p>\tan\theta . \cot\theta = 1\\ \\</p><p>\sin^2\theta+ \cos^2 \theta= 1 \\ \\</p><p>\cosec^2 \theta -\cot^2 \theta = 1 \\ \\</p><p>\sec^2 \theta - \tan^2 \theta = 1......[\text{ii}]$