Question

In the table, trigonometric ratios and the intervals of $\theta$ are given. Match the intervals with the ratios which are positive in those intervals.

$\begin{array}{cc}\theta & \text{gives positive values}\\ \text{p. (0,}\frac{\pi }{2})& \text{1. Only sin}\theta ,\text{cosec}\theta \\ \text{q. (}\frac{\pi }{2},\pi )& \text{2. Only cos}\theta ,\text{sec}\theta \\ \text{r. (}\pi ,\frac{3\pi }{2})& \text{3. Only tan}\theta ,\text{cot}\theta \\ \text{s. (}\frac{3\pi }{2},2\pi )& \text{4. All sin}\theta ,\text{cos}\theta ,\text{tan}\theta ,\text{cot}\theta ,\text{sec}\theta ,\text{cosec}\theta \end{array}$

• A. p - 1, q - 2, r - 3, s - 4
• B. p - 4, q - 3, r - 2, s - 1
• C. p - 4, q - 1, r - 2, s - 3
• D. p - 4, q - 1, r - 3, s - 2

Answer 1

The correct option is D

p - 4, q - 1, r - 3, s - 2

In first quadrant $(0,\frac{\pi }{2})$

We see, both $x$ and $y$ intercepts are positive. $a>0,b>0$

By definition of $\sin \theta$, $\cos \theta$, $\tan \theta$, $\cot \theta$, $\sec \theta$, and $\text{cosec}\theta$

All the trigonometric function gives positive values in first quadrant.

Second Quadrant $(\frac{\pi }{2},\pi )$

We see that $x$ intercept in second quadrant is

negative. Therefore, $a<0,b>0$.

Functions which includes the values of

x - Intercept or adjacent will give negatives values.

All the other functions will give positive values.

By definition we know except $\sin \theta$ & $\text{cosec}\theta$ , all the other functions use adjacent in their trigonometric ratios.

So, only sin$\theta$ & cosec$\theta$ give positive values in second quadrant.

Third quadrant $(\pi ,\frac{3\pi }{2})$

We see both $x$ & $y$ - intercepts are negative

in III quadrant.

Therefore $a<0,b<0$

Both adjacent and opposite are negative.

So, trigonometric ratio which uses both adjacent and opposite gives positive value.

Only $\tan \theta$ & $cot\theta$ use both adjacent and opposite.

In III quadrant, only $\tan \theta$ & $\cot \theta$ give positive values

Remaining trigonometric functions will give negative values.

Fourth quadrant $(\frac{3\pi }{2}$, 2$\pi )$

$x-$intercept is positive while $y-$ intercept negative.

Therefore $a>0,b<0$

Function which uses the values of $y$- intercept gives negative values, remaining functions give positive value.

Only $\cos \theta$ & $\sec \theta$ doesn't will give positive value in fourth quadrant.