Question

Complete the following:


$\tan A$
$\tan B$
$\cot A$
$\cot B$
$\sin A$
$\sin B$
$\cos e c A$
$\cos e c B$
$\cos A$
$\cos B$
$\sec A$
$\sec B$

Accepted Answer

Using the definitions of trigonometric ratios for the given triangle:

$\sin A = \frac{Opposite side}{Hypotenuse} = \frac{B C}{A B} = \frac{3}{5} \Rightarrow \cos e c A = \frac{1}{\sin A} = \frac{5}{3}$ We know,

$\sin B = \frac{A C}{A B} = \frac{4}{5} \Rightarrow \cos e c B = \frac{1}{\sin B} = \frac{5}{4}$

$\cos A = \frac{Adjacent side}{Hypotenuse} = \frac{A C}{A B} = \frac{4}{5} \Rightarrow \sec A = \frac{1}{\cos A} = \frac{5}{4}$

$\cos B = \frac{B C}{A B} = \frac{3}{5} \Rightarrow \sec B = \frac{1}{\cos B} = \frac{5}{3}$

$\tan A = \frac{Opposite side}{Hypotenuse} = \frac{B C}{A C} = \frac{3}{4} \Rightarrow \cot A = \frac{1}{\tan A} = \frac{4}{3}$

$\tan B = \frac{A C}{B C} = \frac{4}{3} \Rightarrow \cot B = \frac{1}{\tan B} = \frac{3}{4}$

$\tan A$ $\frac{3}{4}$

$\tan B$ $\frac{4}{3}$

$\cot A$ $\frac{4}{3}$

$\cot B$ $\frac{3}{4}$

$\sin A$ $\frac{3}{5}$

$\sin B$ $\frac{4}{5}$

$cosec A$ $\frac{5}{3}$

$cosec B$ $\frac{5}{4}$

$\cos A$ $\frac{4}{5}$

$\cos B$ $\frac{3}{5}$

$\sec A$ $\frac{5}{4}$

$\sec B$ $\frac{5}{3}$

O P Malhotra, S K Gupta and, Anubhuti Gangal Solutions for Exercise 1: Exercise

O P Malhotra Mathematics Solutions for Exercise - O P Malhotra, S K Gupta and, Anubhuti Gangal Solutions for Exercise 1: Exercise

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