Question

class 10 maths from trigonometry chapter​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:5 \: cot \theta \: = 3$

$\rm :\longmapsto\:cot\theta = \dfrac{3}{5}$

We know,

$\rm :\longmapsto\:cot\theta = \dfrac{adjacent \: side}{opposite \: side}$

$\rm :\longmapsto\: \dfrac{adjacent \: side}{opposite \: side} = \dfrac{3}{5}$

$\rm :\longmapsto\:adjacent \: side \: = \: 3k$

and

$\rm :\longmapsto\:opposite \: side \: = \: 5k$

We know,

By pythagoras theorem,

$\rm :\longmapsto\: {hypotenuse}^{2} = {opposite \: side}^{2} + {adjacent \: side}^{2}$

$\rm :\longmapsto\: {hypotenuse}^{2} = {(3k)}^{2} + {(5k)}^{2}$

$\rm :\longmapsto\: {hypotenuse}^{2} = {9k}^{2} + {25k}^{2}$

$\rm :\longmapsto\: {hypotenuse}^{2} = {34k}^{2}$

$\rm :\longmapsto\: {hypotenuse}^{} = \sqrt{34} \: {k}$

Now,

$\rm :\longmapsto\:cos\theta = \dfrac{adjacent \: side}{hypotenuse} = \dfrac{3k}{ \sqrt{34} k} = \dfrac{3}{ \sqrt{34} }$

and

$\rm :\longmapsto\:sin\theta = \dfrac{opposite \: side}{hypotenuse} = \dfrac{5k}{ \sqrt{34} k} = \dfrac{5}{ \sqrt{34} }$

Now, Consider

$\rm :\longmapsto\:\dfrac{5sin\theta - 3cos\theta }{4sin\theta + 3cos\theta }$

On substituting the values, evaluated above

$\rm \: = \: \: \dfrac{5 \times \dfrac{5}{ \sqrt{34} } - 3 \times \dfrac{3}{ \sqrt{34} } }{4 \times \dfrac{5}{ \sqrt{34} } + 3 \times \dfrac{3}{ \sqrt{34} } }$

$\rm \: = \: \: \dfrac{\dfrac{25}{ \sqrt{34} } - \dfrac{9}{ \sqrt{34} } }{\dfrac{20}{ \sqrt{34} } + \dfrac{9}{ \sqrt{34} } }$

$\rm \: = \: \: \dfrac{\dfrac{25 - 9}{ \sqrt{34} } }{\dfrac{20 + 9}{ \sqrt{34} } }$

$\rm \: = \: \: \dfrac{\dfrac{16}{ \sqrt{34} } }{\dfrac{29}{ \sqrt{34} } }$

$\rm \: = \: \: \dfrac{16}{29}$

Alternative Method :-

Given that,

$\rm :\longmapsto\:5 \: cot \theta \: = 3$

$\rm :\longmapsto\:cot\theta = \dfrac{3}{5}$

Now, Consider

$\rm :\longmapsto\:\dfrac{5sin\theta - 3cos\theta }{4sin\theta + 3cos\theta }$

$\rm \: = \: \: \dfrac{\dfrac{5sin\theta - 3cos\theta }{sin\theta } }{ \dfrac{4cos\theta + 3sin\theta}{sin\theta } }$

$\red{\bigg \{ \because \: \dfrac{cos\theta }{sin\theta } = cot\theta \bigg \}}$

$\rm \: = \: \: \dfrac{5 - 3cot\theta }{4 + 3cot\theta }$

$\rm \: = \: \: \dfrac{5 - 3 \times \dfrac{3}{5} }{4 + 3 \times \dfrac{3}{5} }$

$\rm \: = \: \: \dfrac{25 - 9}{20 + 9}$

$\rm \: = \: \: \dfrac{16}{29}$