Question

if cosec theta=5/3, find the value of cos theta + tan theta

Answer 1

Hello ,

This problem can be solved by using identities
$\csc( \theta) = \frac{5}{3}$
${ \csc( \theta) }^{2} = { \cot( \theta) }^{2} + 1$

${( \frac{5}{3}) }^{2} = { \cot( \theta) }^{2} + 1$
${ \cot( \theta) }^{2} = \frac{25}{9} - 1$
$\cot( \theta) = \sqrt{ \frac{16}{9} }$
$\cot( \theta) = \frac{4}{3}$
$\cot( \theta) = \frac{1}{ \tan( \theta) }$
$\tan( \theta) = \frac{3}{4}$
1 + tan^2 theta = sec^2 theta
sec^2 theta = 1 + 9/16
= 25/16
sec theta = √25/16
= 5/4
cos theta = 4/5

$\cos( \theta) + \tan( \theta) = \frac{4}{5} \: + \frac{3}{4} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{16 + 15}{20}$
(16 + 15)/20 = 31/20

Answer 2

Answer:

Required value of $\cos(\theta) + \tan(\theta)$is $1 \frac{11}{20}$

Step-by-step explanation:

$\csc(\theta) = \frac{5}{3}$

We know,

${csc}^{2} (\theta) - {cot}^{2} (\theta) = 1$

So,

${cot}^{2} (\theta) = {csc}^{2} (\theta) - 1 \\ {cot}^{2} (\theta) = {( \frac{5}{3}) }^{2} - 1 \\ {cot}^{2} (\theta) = \frac{25}{9} - 1 \\ {cot}^{2} (\theta) = \frac{25 - 9}{9} \\ {cot}^{2} (\theta) = \frac{16}{9} \\ \cot(\theta) = \sqrt{( \frac{16}{9} )} \\ \cot(\theta) = \frac{4}{3}$

We know,

$\tan(\theta) = \frac{1}{ \cot(\theta) } = \frac{1}{ \frac{4}{3} } = \frac{3}{4}$

Again,

${sec}^{2}( \theta) - {tan}^{2} (\theta) = 1 \\ {sec}^{2} (\theta) = {tan}^{2} (\theta) + 1 \\ {sec}^{2} (\theta) = { (\frac{3}{4} )}^{2} + 1 \\ {sec}^{2} (\theta) = \frac{9}{16} + 1 \\ {sec}^{2} (\theta) = \frac{25}{16 \\ } \\ \sec(\theta) = \sqrt{ \frac{25}{16} } \\ \sec(\theta) = \frac{5}{4}$

We know,

$cos(\theta) = \frac{1}{ \sec(\theta) } \\ \cos(\theta) = \frac{1}{ \frac{5}{4} } \\ \cos(\theta) = \frac{5}{4}$

Now,

$\cos(\theta) + \tan(\theta) = \frac{4}{5} + \frac{3}{4} \\ \cos(\theta) + \tan(\theta) = \frac{16 + 15}{20} \\ \cos(\theta) + \tan(\theta) = \frac{31}{20} \\ \cos(\theta) + \tan(\theta) = 1 \frac{11}{20}$