Question

A right angled triangle has one leg of length $1$ unit, another leg of length $x$ units and hypotenuse is of length $y$ units. The angle opposite to the side of length $x$ units is $\theta .$ Then

• A. $y^3–x^3={\sec }^3\theta –{\tan }^3\theta$
• B. $\underset{\theta \to \pi /2}{lim}(y-x)=0$
• C. $\underset{\theta \to \pi /2}{lim}(y^2-x^2)=1$
• D. $\underset{\theta \to \pi /2}{lim}(y^3-x^3)=-1$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $y^3–x^3={\sec }^3\theta –{\tan }^3\theta$
B $\underset{\theta \to \pi /2}{lim}(y-x)=0$
C $\underset{\theta \to \pi /2}{lim}(y^2-x^2)=1$


$x=\tan \theta ,y=\sec \theta$
$y^3–x^3={\sec }^3\theta -{\tan }^3\theta$

$\underset{\theta \to \pi /2}{lim}(\sec \theta -\tan \theta )$
$=\underset{\theta \to \pi /2}{lim}\frac{1-\sin \theta }{\cos \theta }=0$

$\underset{\theta \to \pi /2}{lim}({\sec }^2\theta -{\tan }^2\theta )=1$

$\underset{\theta \to \pi /2}{lim}({\sec }^3\theta -{\tan }^3\theta )$ does not exist.