Question

In fig 8.25, write sec a, tan c, cosec b, cot (90 $-$ c), cos (90 $-$ b) and find the value of sec²a $-$ tan² a.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \sec a=\frac{\mathrm{hypotenuse}}{\mathrm{adjacent}\mathrm{side}\mathrm{of}a}=\frac{\mathrm{SR}}{\mathrm{TR}} \\ \tan \mathit{}c=\frac{\mathrm{opposite}\mathrm{side}\mathrm{of}c}{\mathrm{adjacent}\mathrm{side}\mathrm{of}\mathit{}c}=\frac{\mathrm{SP}}{\mathrm{PT}} \\ \mathrm{cosec}b\mathit{}=\frac{\mathrm{hypotenuse}}{\mathrm{opposite}\mathrm{side}\mathrm{of}\mathit{}b}=\frac{\mathrm{TR}}{\mathrm{TQ}} \end{gathered}$$


In $△$STR, we have:
$\angle$STR + $\angle$SRT + $\angle$TSR = 180^o (Angle sum property of a triangle)
$\Rightarrow$ 90^o + a + $\angle$ TSR = 180^o
$\Rightarrow$ $\angle$ TSR = 90^o − a
Similarly, $\angle$PST = 90^o − c and $\angle$RTQ = 90^o − b

$$\begin{gathered} \mathrm{Now},\mathrm{we}\mathrm{have}: \\ \cot ({90}^{°}-a)=\frac{\mathrm{Adjacent}\mathrm{side}\mathrm{of}(90°-a)}{\mathrm{Opposite}\mathrm{side}\mathrm{of}(90°-a)}=\frac{\mathrm{ST}}{\mathrm{TR}} \\ \sin (90°-c)=\frac{\mathrm{Opposite}\mathrm{side}\mathrm{of}(90°-c)}{\mathrm{Hypotenuse}}=\frac{\mathrm{PT}}{\mathrm{ST}} \\ \cos (90°-b)=\frac{\mathrm{Adjacent}\mathrm{side}\mathrm{of}(90°-b)}{\mathrm{Hypotenuse}}=\frac{\mathrm{TQ}}{\mathrm{TR}} \\ \tan a\mathit{}=\frac{\mathrm{opposite}\mathrm{side}\mathrm{of}a}{\mathrm{adjacent}\mathrm{side}\mathrm{of}a}=\frac{\mathrm{ST}}{\mathrm{TR}} \end{gathered}$$


Now, consider sec² a – tan² a

$$\begin{gathered} ={\left(\frac{\mathrm{SR}}{\mathrm{TR}}\right)}^2-{\left(\frac{\mathrm{ST}}{\mathrm{TR}}\right)}^2 \\ =\frac{{\mathrm{SR}}^2-{\mathrm{ST}}^2}{{\mathrm{TR}}^2} \\ =\frac{{\mathrm{TR}}^2}{{\mathrm{TR}}^2}=1(\mathrm{Pythagoras}\mathrm{theorem}) \end{gathered}$$