If cot A = b/a , Prove that 2sec A +1 / cos A +2 = root of a² + b² / b

Dear Student,

Please find below the solution to the asked query:

Given :  Cot A  = $\frac{\mathrm{b}}{\mathrm{a}}$

We know :  Cot  $\mathrm{\theta }$ =  $\frac{\mathrm{Adjacent}}{\mathrm{Opposite}}$, So Adjacent = b and opposite = a

From Pythagoras theorem we know :

Hypotenuse ² = Adjacent² + Opposite²

Hypotenuse ² = b² + a²

Hypotenuse = $\sqrt{ {\mathrm{a}}^2 + {\mathrm{b}}^2}$

We know :  Cos $\mathrm{\theta }$ =  $\frac{\mathrm{Adjacent}}{\mathrm{Hypotenuse}}$ , So Cos A = $\frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2}}$

And

 Sec $\mathrm{\theta }$ =  $\frac{\mathrm{Hypotenuse}}{\mathrm{Adjacent}}$ , So Sec A = $\frac{\sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2}}{\mathrm{b}}$

So, To show  : $\frac{2 \mathrm{Sec} \mathrm{A}\hspace{0.17em} + 1}{\mathrm{Cos} \mathrm{A} +\hspace{0.17em}2 } = \frac{\sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2}}{\mathrm{b}}$

We take L.H.S.  and substitute values we get

$$\begin{gathered} \frac{2 \mathrm{Sec} \mathrm{A}\hspace{0.17em} + 1}{\mathrm{Cos} \mathrm{A} +\hspace{0.17em}2 } \\ \Rightarrow \frac{2 \left({\displaystyle \frac{\sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2}}{\mathrm{b}}}\hspace{0.17em}\right) + 1}{{\displaystyle \frac{\mathrm{b}}{\sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2}}} +\hspace{0.17em}2 } \\ \Rightarrow \frac{ {\displaystyle \frac{2 \sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2} + b}{\mathrm{b}}}{\displaystyle \hspace{0.17em}} }{{\displaystyle \frac{\mathrm{b} +2 \sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2} }{\sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2}}}} \\ \Rightarrow \frac{2 \sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2} + b}{\mathrm{b}}\times \frac{\sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2} {\displaystyle \hspace{0.17em}} }{{\displaystyle 2 \sqrt{{\mathrm{a}}^2 + {\mathrm{b}}^2} + \mathrm{b} }} \\ \mathbf{\Rightarrow }\mathbf{ }\frac{\sqrt{{\mathbf{a}}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{ }{\mathbf{b}}^{\mathbf{2}}}}{\mathbf{b}} \end{gathered}$$
Hence

L.H.S. =  R.H.S.                                                        ( Hence proved )


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