Question

if 4 tan theta is equal to 3 evaluate 4 sin theta minus cos theta + 1 by 4 sin theta + cos theta minus one​

Answer 1

Appropriate Question :

If $\sf { 4 \tan \theta = 3}$, evaluate $\sf { \dfrac{4\sin \theta - \cos \theta +1 }{ 4\sin \theta + \cos \theta - 1} }$.

Required Solution :

Given that,

• 4 tan θ = 3

So,

$\longmapsto \sf { \tan \theta = \dfrac{3}{4} }$

We know that,

$\longmapsto \sf { \tan \theta = \dfrac{Perpendicular}{Base} }$

So, we get that :

● Perpendicular = 3

● Base = 4

Now, in order to solve the question we need to find the value of sin θ and cos θ. We know that,

» $\longmapsto \sf { \sin \theta = \dfrac{Perpendicular}{Hypotenuse} }$

» $\longmapsto \sf { \cos \theta = \dfrac{Base}{Hypotenuse} }$

Finding hypotenuse :

By pythagoras property,

$\longrightarrow \sf { H^2 = B^2 + P^2 }$

• H = Hypotenuse
• B = Base
• P = Perpendicular

$\longrightarrow \sf { H^2 = 4^2 + 3^2 }$

$\longrightarrow \sf { H^2 = 16 + 9 }$

$\longrightarrow \sf { H^2 = 25 }$

$\longrightarrow \sf { H = \sqrt{25} }$

$\longrightarrow \sf \red{ H = 5 }$

Calculating sin θ :

» $\longmapsto \sf { \sin \theta = \dfrac{Perpendicular}{Hypotenuse} }$

$\longmapsto \sf { \sin \theta = \dfrac{3}{5} }$

Calculating cos θ :

» $\longmapsto \sf { \cos \theta = \dfrac{Base}{Hypotenuse} }$

$\longmapsto \sf { \cos \theta = \dfrac{4}{5} }$

Substituting the values in given expression ,

$\longrightarrow \sf { \dfrac{4\sin \theta - \cos \theta + 1 }{ 4\sin \theta + \cos \theta-1} }$

$\longrightarrow \sf { \dfrac{4 \times \cfrac{3}{5} - \cfrac{4}{5} + 1}{ 4 \times \cfrac{3}{5} + \cfrac{4}{5} -1} }$

$\longrightarrow \sf { \dfrac{\cfrac{4 \times 3}{5} - \cfrac{4}{5} +1 }{ \cfrac{4 \times 3}{5} + \cfrac{4}{5} -1 } }$

$\longrightarrow \sf { \dfrac{\cfrac{12}{5} - \cfrac{4}{5} +1 }{ \cfrac{12}{5} + \cfrac{4}{5} -1} }$

$\longrightarrow \sf { \dfrac{\cfrac{12-4}{5} + 1}{ \cfrac{12+4}{5} - 1} }$

$\longrightarrow \sf { \dfrac{\cfrac{8}{5} +1 }{ \cfrac{16}{5}-1 } }$

$\longrightarrow \sf { \dfrac{\cfrac{8+5}{5} }{ \cfrac{16-5}{5} } }$

$\longrightarrow \sf { \dfrac{\cfrac{13}{5} }{ \cfrac{11}{5} } }$

$\longrightarrow \sf { \dfrac{13}{\not{5}} \times \dfrac{\not {5}}{11} }$

$\longrightarrow\boxed{ \sf \red { \dfrac{13}{11} } }$

Therefore,

$\sf { \dfrac{4\sin \theta - \cos \theta +1 }{ 4\sin \theta + \cos \theta - 1} = \sf \red { \dfrac{13}{11}} }$

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Know More !

» $\longmapsto \sf { \sin \theta = \dfrac{Perpendicular}{Hypotenuse} }$

» $\longmapsto \sf { \cosec \theta = \dfrac{Hypotenuse}{Perpendicular} }$

» $\longmapsto \sf { \cos \theta = \dfrac{Base}{Hypotenuse} }$

» $\longmapsto \sf { \sec \theta = \dfrac{Hypotenuse}{Base} }$

» $\longmapsto \sf { \tan \theta = \dfrac{Perpendicular}{Base} }$

» $\longmapsto \sf { \cot \theta = \dfrac{Base}{Perpendicular} }$