Question

If 3 sin theta -4 cos theta =0,find the values of tan theta ,sec theta and cosec theta..

Hey guys plz ans.

Spam ans. would be reported ...
^_^​

Answer 1

Answer:

tan theta =4/5

sec theta = 5/3

cosec theta =5/4

Answer 2

Answer:

$\huge\mathcal{\green{Bonjour!}}$

$\huge\mathfrak{\red{Answer}}$

$\huge\fbox{\purple{tan \ x \ = 4/3}}$

$\huge\fbox{\purple{sec \ x \ = \ 5/3 }}$

$\huge\fbox{\purple{cosec \ x \ = 5/4}}$

Step-by-step explanation:

☆ Question:-

If 3 sin theta -4 cos theta = 0, then find the values of tan theta ,sec theta and cosec theta.

☆ Given:-

• 3 sin x - 4 cos x = 0

☆ To find:-

• tan x

• sec x

• cosec x

☆ Required Solution:-

Given that 3 sin x - 4 cos x = 0

: ➝ 3 sin x = 4 cos x. [ by moving (- 4 cos x) to RHS]

$= > \frac{sin \: x}{cos \: x} = \frac{4}{3}$

$= > tan \: x = \frac{4}{3}$

[As tan x = sin x/cos x]

$\huge\mathcal{\green{Now,}}$

We have

: ➝ 1 + tan^2x = sec^2 x

$= > \: {sec}^{2} x = 1 + ({ \frac{4}{3} )}^{2}$

$= > {sec}^{2} x = 1 + \frac{16}{9}$

$= > {sec}^{2} x = \frac{9 + 16}{9} = \frac{25}{9}$

$= > sec \: x = \sqrt{ \frac{25}{9} }$

$= > sec \: x = \frac{5}{3}$

$\huge\mathcal{\green{Now,}}$

$cot \: x = \frac{1}{tan \: x}$

$\huge\mathcal{\green{Therefore,}}$

$cot \: x = \frac{3}{4}$

$\huge\mathcal{\green{Now,}}$

We have

: ➝ cosec^2x = cot^2 x + 1

$= > {cosec}^{2} x = ( { \frac{3}{4} )}^{2} + 1$

$= > {cosec }^{2} x = \frac{9}{16} + 1$

$= > {cosec}^{2} x = \frac{9 + 16}{16} = \frac{25}{16}$

$= > cosec \: x = \sqrt{ \frac{25}{16} } = \frac{5}{4}$

$\huge\mathcal{\green{All \ the \ very \ best!}}$

$\huge\mathfrak{\red{@MissTranquil}}$

$\huge\fbox{\orange{be \ brainly}}$

____________________________________________

☆ Note:-

• Here, theta is taken as x .