Question

For acute angle $\theta$, find ${\text{cosec}}^2\theta -{\cot }^2\theta$

• A. $2$
• B. $3$
• C. $0$
• D. $1$

Answer 1

Answer: (D)

$1$

Consider right angled $△ABC,\angle B={90}^{\circ },\angle A=\theta$
Also, $AB=x,BC=y$ and $AC=r$

$x^2+y^2=r^2$ ...By Pythagoras theorem

$\therefore {cosec}^2\theta -{\cot }^2\theta =(\frac{1}{{\sin }^2\theta }-{\cot }^2\theta )$

$=(\frac{1}{{\left(\frac{y}{r}\right)}^2}-\frac{x^2}{y^2})=(\frac{r^2}{y^2}-\frac{x^2}{y^2})$

$=\frac{r^2-x^2}{y^2}=\frac{y^2}{y^2}=1$.