Question

If sin $\theta$ = $\frac{45}{53}$, find the value of cosec² $\theta$ $-$ cot² $\theta$

Answer 1

Given:
$\sin \theta =\frac{45}{53}$
We know that $\sin \mathrm{\theta }=\frac{\mathrm{perpendicular}}{\mathrm{hypotenuse}}=\frac{45}{53}$
Let perpendicular = 45k and hypotenuse = 53k, where k is a positive constant
In the right triangle, we have:

$$\begin{gathered} {\mathrm{Hypotenuse}}^2={\mathrm{perpendicular}}^2+{\mathrm{base}}^2 \\ \Rightarrow {\left(53k\right)}^2={\left(45k\right)}^2+{\mathrm{base}}^2 \\ \Rightarrow 2809k^2=2025k^2+{\mathrm{base}}^2 \\ \Rightarrow \mathrm{Bas}{e}^2=784k^2 \\ \Rightarrow \mathrm{Base}=\pm 28k \\ \Rightarrow \mathrm{Base}=+28k(\because \mathrm{Length}\mathrm{cannot}\mathrm{be}\mathrm{negative}) \\ \mathrm{Now},\mathrm{we}\mathrm{have}: \\ \mathrm{cosec}\mathrm{\theta }=\frac{\mathrm{hypotenuse}}{\mathrm{perpendicular}}=\frac{53}{45} \\ \cot \mathrm{\theta }=\frac{\mathrm{Base}}{\mathrm{Perpendicular}}=\frac{28}{45} \end{gathered}$$

$$\begin{gathered} \mathrm{Now},{\mathrm{cosec}}^2\mathrm{\theta }-{\cot }^2\mathrm{\theta }={\left(\frac{53}{45}\right)}^2-{\left(\frac{28}{45}\right)}^2 \\ =\frac{{53}^2-{28}^2}{{45}^2}=\frac{2025}{2025}=1 \end{gathered}$$