Question

Q)32 If 4 cot x = 3, then find the value of
sin x +cOS X upon sin x +cos x​

Answer 1

${\huge{\rm{\underline{\underline{Question:-}}}}}$

Q) If 4cot x = 3 , then find the value of

$\sf \: \frac{sin \: x - cos \: x}{sin \: x \: + cos \: x}$

${\huge{\underline{\underline{\rm{Answer\checkmark}}}}}$

☆ Given :

• 4cot x = 3

☆ To Find :

• $\rm \: \frac{sin \: x - cos \: x}{sin \: x + cos \: x}$

☆ Solution :

$\rm \: 4cot \: x = 3$

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

$\mapsto \rm \: \frac{base}{perpendicular} = \frac{3}{4}$

Consider a Right angled triangle in which

• base = 3a
• perpendicular = 4a

Using Pythagoras Theorem -

$\sf \: hypotenuse {}^{2} = base { }^{2} + perpendicular {}^{2}$

$\sf\leadsto \: hypotenuse {}^{2} = {(3a)}^{2} + {(4a)}^{2}$

$\leadsto \sf \: hypotenuse {}^{2} = 9 {a}^{2} + 16 {a}^{2}$

$\leadsto \sf \: hypotenuse = \sqrt{25 {a}^{2} }$

$\leadsto{ \boxed{ \sf \: hypotenuse = 5a \: }}$

So ,

$\sf \leadsto sin \: x = \frac{perpendicular}{hypotanuse}$

$\leadsto \sf \: sin \: x = \frac{4 \cancel{a}}{5 \cancel{a}}$

$\leadsto \underline{ \boxed{ \sf \: sin \: x = \frac{4}{5} \: }}$

And ,

$\leadsto \sf \: cos \: x = \frac{base}{hypotenuse}$

$\leadsto \sf \: cos \: x = \frac{3 \cancel{a}}{5 \cancel{a}}$

$\leadsto \underline{ \boxed{ \sf \: cos \: x = \frac{3}{5} \: }}$

Now , Just put the values in the term which we want to simplify .

$\rm \mapsto \: \frac{sin \: x - cos \: x}{sin \: x + cos \: x}$

$\mapsto \rm \: \frac{ \frac{4}{5} - \frac{3}{5} }{ \frac{4}{5} + \frac{3}{5} }$

$\mapsto \: \frac{ \frac{4 - 3}{ \cancel5} }{ \frac{4 + 3}{ \cancel5} }$

$\mapsto \: \frac{1}{7}$

So ,

$\green{\boxed{ \rm \: \frac{sin \: x - cos \: x}{sin \: x + cos \: x} = \frac{1}{7}} }$

_________________________

★ Formulas Used :

• $\tt \: cot \: x = \frac{base}{perpendicular}$
• $\tt \: hypotenuse {}^{2} = perpendicular {}^{2} + base {}^{2}$
• $\tt \: sin \: \theta = \frac{perpendicular}{hypotenuse}$
• $\tt \: cos \: \theta = \frac{base}{hypotenuse}$

★ Know More :

• $\tt \: tan \: \theta = \frac{perpendicular}{base}$
• $\tt \: {sin}^{2} \: \theta + {cos}^{2} \: \theta = 1$
• $\tt \: {tan}^{2} \: \theta + 1 = {sec}^{2} \: \theta$
• $\tt \: 1 + {cot}^{2} \: \theta = {cosec}^{2} \: \theta$

Answer 2

the above answer is absolutely correct