Question

if cosec theta =13/12, find the value of 2 sin theta - 3 cos theta / 4 sin theta - 9 cos theta​

Answer 1

Solution :-

Given ,

• cosecθ = $\sf\dfrac{13}{12}$= $\sf\dfrac{Hypotenuse}{opposite}$

To find ,

$\qquad\bullet \;\dfrac{2\sin\theta-3\cos\theta}{4\sin\theta-9\cos\theta}$

Taking a right angled triangle ABC with ,

• AB = Height = 12
• BC = Base
• AC = Hypotenuse = 13

So , using Pythagoras theorem

AC² = AB² + BC²

13² = 12² + BC²

BC = $\sf\sqrt{169-144}$

BC = $\sf\sqrt{25}$

BC = 5

Now ,

• sinθ = $\sf\dfrac{12}{13}$

• cosθ = $\sf \dfrac{5}{13}$

Now , substituting

$\longrightarrow \sf\dfrac{2sin\theta-3cos\theta}{4sin\theta-9cos\theta}$

$\displaystyle\sf\longrightarrow \dfrac{2\bigg(\dfrac{12}{13}\bigg)-3\bigg(\dfrac{5}{13}\bigg)}{4\bigg(\dfrac{12}{13}\bigg)-9\bigg(\dfrac{5}{13}\bigg)}$

$\displaystyle\sf\longrightarrow \dfrac{\bigg(\dfrac{24}{13}\bigg)-\bigg(\dfrac{15}{13}\bigg)}{\bigg(\dfrac{48}{13}\bigg)-\bigg(\dfrac{45}{13}\bigg)}$

$\displaystyle\sf\longrightarrow \dfrac{\bigg(\dfrac{24 - 15}{13}\bigg)}{\bigg(\dfrac{48 - 45}{13}\bigg)}$

$\displaystyle\sf\longrightarrow \dfrac{\bigg(\dfrac{9}{13}\bigg)}{\bigg(\dfrac{3}{13}\bigg)} = \dfrac{9}{ 3}$

$\displaystyle\sf\longrightarrow \underline{\boxed{\textbf{\textsf{3}}}}$

$\therefore \underline{\underline{\sf\orange{Hence, \;\dfrac{2sin\theta-3cos\theta}{4sin\theta-9cos\theta}=3}}}$

Answer 2

$cosecθ = \sf\dfrac{13}{12}$

$= \sf\dfrac{Hypotenuse}{opposite}$

To find ,

$\qquad\bullet \;\dfrac{2\sin\theta-3\cos\theta}{4\sin\theta-9\cos\theta}$

Taking a right angled triangle ABC with ,

AB = Height = 12

BC = Base

AC = Hypotenuse = 13

So , using Pythagoras theorem

AC² = AB² + BC²

13² = 12² + BC²

$BC = \sf\sqrt{169-144}$

$BC = \sf\sqrt{25}$

BC = 5

$sinθ = \sf\dfrac{12}{13}$

$cosθ = \sf \dfrac{5}{13}$

Then,

$\longrightarrow \sf\dfrac{2sin\theta-3cos\theta}{4sin\theta-9cos\theta}$

$\displaystyle\sf\longrightarrow \dfrac{2\bigg(\dfrac{12}{13}\bigg)-3\bigg(\dfrac{5}{13}\bigg)}{4\bigg(\dfrac{12}{13}\bigg)-9\bigg(\dfrac{5}{13}\bigg)}$

$\displaystyle\sf\longrightarrow \dfrac{\bigg(\dfrac{24}{13}\bigg)-\bigg(\dfrac{15}{13}\bigg)}{\bigg(\dfrac{48}{13}\bigg)-\bigg(\dfrac{45}{13}\bigg)}$

$\displaystyle\sf\longrightarrow \dfrac{\bigg(\dfrac{24 - 15}{13}\bigg)}{\bigg(\dfrac{48 - 45}{13}\bigg)}$

$\displaystyle\sf\longrightarrow \dfrac{\bigg(\dfrac{9}{13}\bigg)}{\bigg(\dfrac{3}{13}\bigg)} = \dfrac{9}{ 3}$

$\displaystyle\sf\longrightarrow \underline{\boxed{\textbf{\textsf{3}}}}$

${\underline{\sf\pink{Hope \: it \: is \: helpful}}}$