Question

If 4 cotΘ =3, then sinΘ - cosΘ/ sinΘ + cosΘ is equal to
(a) 5/6
(b) 1/7
(c) 6/7
(d) 3/4

Answer 1

Given :-

$\quad \leadsto \quad \sf 4 \cot \theta = 3$

To Find :-

$\quad \qquad \dfrac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta}$

Solution :-

Consider given ;

$\quad \leadsto \quad \sf 4 \cot \theta = 3$

${ : \implies \quad \sf \cot \theta = \dfrac{3}{4} \quad \qquad ----( i ) }$

Now , consider what we have to find ;

$\quad \leadsto \quad \dfrac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta}$

Dividing both numerator and denominator by $\sin \theta$ we have ;

${ : \implies \quad \sf \dfrac{\dfrac{\sin \theta - \cos \theta}{\sin \theta}}{\dfrac{\sin \theta + \cos \theta}{\sin \theta}}}$

Seperating The LCM of numerator and denominator we have ;

${ : \implies \quad \sf \dfrac{ \dfrac{\sin \theta}{\sin \theta} - \dfrac{\cos \theta}{\sin \theta}}{\dfrac{\sin \theta}{\sin \theta} + \dfrac{\cos \theta}{\sin \theta}} }$

${ : \implies \quad \sf \dfrac{ \dfrac{\cancel{\sin \theta}}{\cancel{\sin \theta}} - \dfrac{\cos \theta}{\sin \theta}}{\dfrac{\cancel{\sin \theta}}{\cancel{\sin \theta}} + \dfrac{\cos \theta}{\sin \theta}} }$

Now , we have ;

${ : \implies \quad \sf \dfrac{ 1 - \dfrac{\cos \theta}{\sin \theta}}{1 + \dfrac{\cos \theta}{\sin \theta}} }$

We knows that ;

$\quad \qquad { \bigstar { \underline { \boxed { \red { \bf { \dfrac{\cos \theta}{\sin \theta} = \cot \theta }}}}}}{\bigstar}$

Using this we have ;

${ : \implies \quad \sf \dfrac{ 1 - \cot \theta }{1 + \cot \theta } }$

Now , putting the values we have obtained in eq. ( i ) we get ;

${ : \implies \quad \sf \dfrac{ 1 - \dfrac{3}{4} }{1 + \dfrac{3}{4} } }$

${ : \implies \quad \sf \dfrac{\dfrac{4-3}{4}}{\dfrac{4+3}{4}}}$

${ : \implies \quad \sf \dfrac{\dfrac{1}{\cancel{4}}}{\dfrac{7}{\cancel{4}}}}$

${ : \implies \quad \tt \dfrac{1}{7}}$

$\quad \qquad { \bigstar { \underline { \boxed { \red { \underbrace { \bf \pmb { \dfrac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} = \dfrac{1}{7} }}}}}}}{\bigstar}$

Henceforth , The Required Answer is option ( b ) $\sf \dfrac{1}{7}$

Answer 2

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$\maltese\; {\underline{\underline {\textsf{\textbf{Question :}}}}}$

• If 4 cot ∅ =3, then sin∅ - cos∅ / sin∅+ cos∅ is equal to

$\maltese\; {\underline{\underline {\textsf{\textbf{To Find :}}}}}$

$\bigstar \; {\underline{\boxed{\bf{ \frac{\sin \theta - \cos \theta }{\sin \theta + \cos \theta} }}}}$

$\maltese\; {\underline{\underline {\textsf{\textbf{Required Answer :}}}}}$

• The value of the expression equals to the fraction 1/7 [ option ( B ) ]

$\maltese\; {\underline{\underline {\textsf{\textbf{Full Solution :}}}}}$

★ Now we have the been provided with the value of 4 cot ∅ which is equivalent  to 3

• $\sf 4 \; cot \theta = 3$

• $\sf cot \theta = 3/4 ---- ( 1 )$  

${\bigstar \; {\underline{\pmb{\sf{ Since , it's \;known \; that\; trignometric \; ratios \; are \; as \; follows :}}}}}$

$\boxed{\begin{aligned}&\bullet\:\sin\theta = \rm{Opposite\:Side/Hypotenuse }\\&\bullet\:\cos\theta = \rm{Adjacent\:Side/Hypotenuse}\\&\bullet\:\tan\theta= \rm{Opposite\:Side/Adjacent\:Side }\\&\bullet\:\csc\theta =\rm{Hypotenuse/Opposite\:Side }\\&\bullet\:\sec\theta =\rm{Hypotenuse/Adjacent\:Side }\\&\bullet\:\cot\theta =\rm{Adjacent\:Side/Opposite\:Side }\\\end{aligned}}$  

★ From the above info :

• we have the values of the adjacent and the opposite sides of the triangle now let's find out the value of hypotenuse of the triangle

* Pythagoras Theorem :  

Pythagoras theorem states the relation between the sides of a right angled triangle where the square of its hypotenuse equals to the sums of squares of its other two sides

➽ Using the formula ,

$\bigstar \; {\underline{\boxed{\bf{ Hypotenuse^2 = Opposite \; side^2 + Adjecent \; side ^2 }}}}$

➞ Now let's find out the value of the hypotenuse of the triangle

${ \longrightarrow } \rm ( Hypotenuse ) ^2 = ( Adjacent \; side ) ^2 + ( Opposite \; side ) ^2$

${ \longrightarrow } \rm ( Hypotenuse ) = \sqrt{( Adjacent \; side ) ^2 + ( Opposite \; side ) ^2}$

${ \longrightarrow } \rm ( Hypotenuse ) = \sqrt{9 + 16}$

${ \longrightarrow } \rm ( Hypotenuse ) = \sqrt{25}$

${ \longrightarrow } \rm {\red{\underline{\underline{( Hypotenuse ) = 5 \; units}}}}$

~ Therefore the value of the hypotenuse of the triangle is 5 units

$\maltese\; {\underline{\underline {\textsf{\textbf{Diagram of triangle :}}}}}$

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1) \put(3.4,3.5) {\large \bf 5 } \put(.3,2.5){\large\bf 4}\put(2.8,.3){\large\bf 3}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf \theta }\end{picture}$  

★ Since we know that the sin ratio is

$\rightarrow \rm \sin \theta = Opposite \; side / hypotenuse$

$\rightarrow \rm \sin \theta = 4/5$

★ Since we know that the cos ratio is

$\rightarrow \rm \cos \theta = Adjacent \; side / hypotenuse$

$\rightarrow \rm \cos \theta = 3/5$

─  Now let's substitute the value of the functions which were figured out

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

$\longrightarrow \sf \dfrac{4/5 - 3/5}{4/5 + 3/5}$

$\longrightarrow \sf \dfrac{1/5}{7/5}$

$\longrightarrow \sf 1/7$

$\bf --------- - - - Alternate \; Method --------- - - - -$

• We know that $\rm cosec^2 \theta - cot^2 \theta = 1$

$\longrightarrow \rm cosec^2 \theta = 1 + cot^2 \theta \cdots \cdots \tt transposing$

$\longrightarrow \rm cosec^2 \theta = 1 + ( 3/4 ) ^2$

$\longrightarrow \rm cosec^2 \theta = 1 + 9/16$

$\longrightarrow \rm cosec^2 \theta = 16 /16+ 9/16$

$\longrightarrow \rm cosec^2 \theta = 25/16$

$\longrightarrow \rm cosec \theta = 5/4$

• From the table above

★ We can state that sin function is the reciprocal of cosec

$\longrightarrow \rm cosec \theta = 5/4$

$\longrightarrow \rm sin \theta = 4/5$

• We know that $\rm sin^2 \theta + cos^2 \theta = 1$  

$\longrightarrow \rm cos^2 \theta = 1 - sin^2 \theta \cdots \cdots \tt transposing$

$\longrightarrow \rm cos^2 \theta = 1 - (4/5)^2$

$\longrightarrow \rm cos^2 \theta = 1 - 16/25$

$\longrightarrow \rm cos^2 \theta = 25/ 25 - 16/25$

$\longrightarrow \rm cos^2 \theta = 9/25$

$\longrightarrow \rm cos \theta = 3/5$

• Evaluating the above expression

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

$\longrightarrow \sf \dfrac{4/5 - 3/5}{4/5 + 3/5}$

$\longrightarrow \sf \dfrac{1/5}{7/5}$

$\longrightarrow \sf 1/7$

$\maltese\; {\underline{\underline {\textsf{\textbf{Therefore :}}}}}$

• The value of the above given expression is 1/7

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