91Ó°ÊÓ

Here we have to prove the following reciprocal identities :-

(a) $csc\mathrm{θ}=\frac{1}{\sin \mathrm{θ}}$

(b) $sec\mathrm{θ}=\frac{1}{\cos \mathrm{θ}}$

(c) $cot\mathrm{θ}=\frac{1}{\tan \mathrm{θ}}$

We will apply trigonometry ratios, to prove these reciprocal identities.

Consider the following right angled triangle.

For this right angled triangle, sine function is defined as perpendicular upon hypotenuse.

That is :-

$\sin \mathrm{θ}=\frac{AB}{AC}$ $.......\left(1\right)$

and cosecant function is defined as hypotenuse upon perpendicular.

That is :-

$csc\mathrm{θ}=\frac{AC}{AB}$ $..........\left(2\right)$

From $\left(1\right)$, we have :-

$\sin \mathrm{θ}=\frac{AB}{AC}$

Take reciprocals on both sides, then we have :-

$\frac{1}{\sin \mathrm{θ}}=\frac{AC}{AB}$

By comparing $\left(2\right)and\left(3\right)$, we have :-

The right hand sides of both equations are equal, so left hand sides are also equal. This gives us :-

$csc\mathrm{θ}=\frac{1}{\sin \mathrm{θ}}$

Hence proved.

Consider the following right angled triangle.

For this right angled triangle, cosine function is defined as base upon hypotenuse.

That is :-

$\cos \mathrm{θ}=\frac{BC}{AC}$$..........\left(4\right)$

and secant function is defined as hypotenuse upon base.

That is :-

$sec\mathrm{θ}=\frac{AC}{BC}$$..........\left(5\right)$

From $\left(4\right)$, we have :-

$\cos \mathrm{θ}=\frac{BC}{AC}$

Take reciprocals on both sides, then we have :-

$\frac{1}{\cos \mathrm{θ}}=\frac{AC}{BC}$$.........\left(6\right)$

By comparing $\left(5\right)and\left(6\right)$, we have :-

The right hand sides of both equations are equal, so left hand sides are also equal. This gives us :-

$sec\mathrm{θ}=\frac{1}{\cos \mathrm{θ}}$

Hence proved.

Consider the following right angled triangle.

For this right angled triangle, tangent function is defined as perpendicular upon base.

That is :-

$\tan \mathrm{θ}=\frac{AB}{BC}$$........\left(7\right)$

and cotangent function is defined as base upon perpendicular.

That is :-

$cot\mathrm{θ}=\frac{BC}{AB}$ $.........\left(8\right)$

From $\left(7\right)$, we have :-

$\tan \mathrm{θ}=\frac{AB}{BC}$

Take reciprocals on both sides, then we have :-

$\frac{1}{\tan \mathrm{θ}}=\frac{BC}{AB}$$.........\left(9\right)$

By comparing $\left(8\right)and\left(9\right)$, we have :-

The right hand sides of both equations are equal, so left hand sides are also equal. This gives us :-

$cot\mathrm{θ}=\frac{1}{\tan \mathrm{θ}}$

Hence proved.