Question

Show that: $\frac{\mathrm{sinA}}{(\mathrm{secA}+\mathrm{tanA}-1)}+\frac{\mathrm{cosA}}{(\mathrm{cosecA}+\mathrm{cotA}-1)}=1.$

Answer 1

To prove :

$\frac{\mathrm{sinA}}{(\mathrm{secA}+\mathrm{tanA}-1)}+\frac{\mathrm{cosA}}{(\mathrm{cosecA}+\mathrm{cotA}-1)}=1$

$\mathrm{L}.\mathrm{H}.\mathrm{S}=$$\frac{\mathrm{sinA}}{(\mathrm{secA}+\mathrm{tanA}-1)}+\frac{\mathrm{cosA}}{(\mathrm{cosecA}+\mathrm{cotA}-1)}$

Let us express $\sec \mathrm{A}$, $\mathrm{cosec}\mathrm{A}$and $\cot \mathrm{A}$in terms of $\mathrm{sinA}$ and $\mathrm{cosA}$, we get

$=\frac{\mathrm{sinA}}{\left[\left(\frac{1}{\cos \mathrm{A}}\right)+\left(\frac{\mathrm{sinA}}{\mathrm{cosA}}\right)–1\right]}+\frac{\mathrm{cosA}}{\left[\left(\frac{1}{\sin \mathrm{A}}\right)+\left(\frac{\mathrm{cosA}}{\mathrm{sinA}}\right)–1\right]}$

$=\frac{\mathrm{sinA}}{\left[\frac{1+\mathrm{sinA}}{\cos \mathrm{A}}–1\right]}+\frac{\mathrm{cosA}}{\left[\frac{1+\mathrm{cosA}}{\sin \mathrm{A}}–1\right]}$

$=\frac{\mathrm{sinA}}{\left[\frac{1+\mathrm{sinA}-\cos A}{\cos \mathrm{A}}\right]}+\frac{\mathrm{cosA}}{\left[\frac{1+\mathrm{cosA}-\sin A}{\sin \mathrm{A}}\right]}$

$=\frac{\mathrm{sinAcos}\mathrm{A}}{\left(1+\mathrm{sinA}-\mathrm{cosA}\right)}+\frac{\mathrm{cosAsin}\mathrm{A}}{\left(1+\mathrm{cosA}-\mathrm{sinA}\right)}$

$=\frac{\mathrm{sinAcos}\mathrm{A}}{1+(\mathrm{sinA}-\mathrm{cosA})}+\frac{\mathrm{cosAsin}\mathrm{A}}{1-(\mathrm{sinA}-\mathrm{cosA})}$ [Taking negative sign common from$(\mathrm{cosA}-\mathrm{sinA})$ term of the denominator of the second fraction]

$=\frac{\mathrm{sinAcos}\mathrm{A}\left[1-(\mathrm{sinA}-\mathrm{cosA})\right]+\mathrm{cosAsin}\mathrm{A}\left[1+(\mathrm{sinA}-\mathrm{cosA})\right]}{1^2-{(\mathrm{sinA}-\mathrm{cosA})}^2}$ $\left[\because {\mathrm{a}}^2-{\mathrm{b}}^2=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})\right]$

$=\frac{\mathrm{sinAcos}\mathrm{A}-{\sin }^2\mathrm{AcosA}+{\mathrm{sinAcos}}^2\mathrm{A}+\mathrm{cosAsin}\mathrm{A}+{\mathrm{cosAsin}}^2\mathrm{A}-{\cos }^2\mathrm{AsinA}}{1^2-\left\{{\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}-2\mathrm{sinAcosA}\right\}}$ (like terms with opposite sign cancel each other)

$=\frac{2\mathrm{sinAcos}\mathrm{A}}{1-1+2\mathrm{sinAcosA}}$ $[\because {\sin }^2\mathrm{A}+{\cos }^2\mathrm{A}=1]$

$=\frac{2\mathrm{sinAcos}\mathrm{A}}{2\mathrm{sinAcosA}}$

$=1$

$=\mathrm{R}.\mathrm{H}.\mathrm{S}$

$\because \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}$

$\therefore$$\frac{\mathrm{sinA}}{(\mathrm{secA}+\mathrm{tanA}-1)}+\frac{\mathrm{cosA}}{(\mathrm{cosecA}+\mathrm{cotA}-1)}=1$

Hence, proved.