Question

Show that :
$\sf{\sqrt{\dfrac{1+SinA}{1-SinA}}=Sec A + tanA }$

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Answer 1

AnswEr-:

• Show that -:

• $\bf{\mathrm{\sqrt{\dfrac{1+SinA}{1-SinA}}=Sec A + tanA }}$

Now ,

• Solving L.H.S -:

• $\bf{\mathrm{L.H.S = \sqrt{ \dfrac{1+SinA}{1-SinA}}}}$

• $\bf{\mathrm{R.H.S = Sec A + tan A}}$

• Now Solving L.H.S -:

• $\bf{\mathrm{L.H.S = \sqrt{ \dfrac{1+SinA}{1-SinA}}}}$

• $\sf{\bf{ By \: Multiplying \:by \:their \: Conjugates \: in\:the L.H.S }}$

• $\longrightarrow{\bf{\mathrm{\sqrt{ \dfrac{1+SinA \times 1 + Sin A }{1-SinA \times 1 + Sin A}}}}}$

• $\longrightarrow{\bf{\mathrm{\sqrt{ \dfrac{(1+SinA)^{2} }{1-SinA \times 1 + Sin A}}}}}$

• $\longrightarrow{\bf{\mathrm{\sqrt{ \dfrac{ (1+SinA)^{2} }{(1-SinA ) (1 + Sin A)}}}}}$

• $\longrightarrow{\bf{\mathrm{\sqrt{ \dfrac{ (1+SinA)^{2} }{ 1 - Sin^{2} A }}}}}$

• $\sf{\star {As \: we \:know \: that -: \:\: 1 - Sin^{2} A = cos^{2} A }}$

• $\longrightarrow{\bf{\mathrm{\sqrt{ \dfrac{ (1+SinA)^{2} }{ cos^{2} A }}}}}$

• $\longrightarrow{\bf{\mathrm{\sqrt{ \left(\dfrac{1+SinA }{ cos A }\right)^{2}}}}}$

• $\sf{\star {As \: we \:know \: that -: \:\: \bf{\sqrt{(y)^{2}} = y }}}$

• $\longrightarrow{\bf{\mathrm{\sqrt{ \left(\dfrac{1+SinA }{ CosA }\right)^{2}}}}}$

• $\longrightarrow{\bf{\mathrm{ \dfrac{1+SinA }{ cos A }}}}$

• $\longrightarrow{\bf{\mathrm{ \dfrac{1 }{ cos A } + \dfrac{SinA }{ cos A }}}}$

• $\sf{\star {As \: we \:know \: that -: \:\: \dfrac{1}{cosA} = Sec A }}$

• $\longrightarrow{\bf{\mathrm{ Sec A + \dfrac{SinA }{ cos A }}}}$

• $\sf{\star {As \: we \:know \: that -: \:\: \dfrac{SinA}{cosA} = tan A }}$

• $\longrightarrow{\bf{\mathrm{ Sec A + tanA }}}$

Therefore ,

• $\longrightarrow{\bf{\mathrm{ L.H.S = Sec A + tanA }}}$

• and ,

• $\longrightarrow{\bf{\mathrm{ R.H.S = Sec A + tanA }}}$

• From this we can see that ,

• $\longrightarrow{\bf{\mathrm{ (L.H.S = Sec A + tanA) = (R.H.S = Sec A + tanA) }}}$

Therefore,

• $\longrightarrow{\bf{\mathrm{ L.H.S = R.H.S }}}$

• $\longrightarrow{\bf{\mathrm{ Hence,\: Verified!}}}$

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