Question

show that ✓1+sinA/1-sina =secA+tanA (O<0<90)​

Answer 1

Question :

Show that √[(1 + sinA)/(1 - sinA)] = secA + tanA

Solution :

Starting from LHS :

$\longmapsto\displaystyle\sf \sqrt{ \dfrac{1 + sinA}{1 - sinA}} \\\\ \\ \longmapsto\sf \sqrt{ \dfrac{(1 + sinA) (1 + sinA)}{(1 - sinA) (1 + sinA)} } \qquad\quad \bigg[Rationalising\ the \ denominator \bigg] \\\\ \\ \longmapsto\sf \sqrt{ \dfrac{ (1 + sinA)^2}{ 1 - sin^2 A}} \\\\ \\ \longmapsto\sf \sqrt{ \dfrac{(1 + sinA)^2 }{cos^2 A}} \qquad\quad \bigg[\because cos^2 A = 1 - sin^2 A \bigg] \\\\ \\ \longmapsto\sf \dfrac{ 1 + sinA}{cosA} \\\\ \\ \longmapsto\sf \dfrac{1}{cosA} + \dfrac{sinA}{cosA}$

$\sf Identities : \underline{\underline{\bold{ \dfrac{1}{cosA} = secA}}} \ \ \ ; \ \ \underline{\underline{\bold{\dfrac{sinA}{cosA} = tanA}}}$

$\longmapsto\underline{\large{\boxed{\bold{secA + tanA}}}} \ \bigstar{\red}$

$\therefore \ \boldsymbol{ \sqrt{ \dfrac{1 + sinA}{1 - sinA}} = secA + tanA \qquad[Showed]}$

Answer 2

Question:

show that

√1+sinA/1-sina =secA+tanA (O<0<90)

Solution:

L.H.S =>

$= \sqrt{ \frac{1 + \sin(a) }{1 - \sin(a) } }$

$= \sqrt{ \frac{1 + \sin(a) }{1 - \sin(a) } \times \frac{1 + \sin(a) }{1 + \sin(a) } }$

$= \sqrt{ \frac{ {(1 + \sin(a)) }^{2} }{(1 + \sin(a) )(1 - \sin(a) )} }$

$= \sqrt{ \frac{ {(1 + \sin(a)) }^{2} }{1 - { \sin }^{2}(a) } }$

$= \sqrt{ \frac{ {(1 + \sin(a)) }^{2} }{ { \cos}^{2} (a) } }$

$=\frac{ (1 + sin(a))}{cos(a)}$

$= \frac{1}{cos(a)} + \frac{sin(a)}{cos(a)}$

$= \sec(a) + \tan(a)$

=> R.H.S

proved