Question

Q1.if cosec theta =13/12, find the value of 2 sin theta - 3 cos theta / 4 sin theta - 9 cos theta​

Answer 1

Step-by-step explanation:

Solution :-

Given ,

$cosecθ = \sf\dfrac{13}{12} </p><p> \ = \sf\dfrac{Hypotenuse}{opposite} </p><p>$

To find ,

$\qquad\bullet \;\dfrac{2\sin\theta-3\cos\theta}{4\sin\theta-9\cos\theta}∙</p><p>$

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

$BC = \sf\sqrt{169-144} </p><p></p><p>$

BC = √25

BC = 5

Now ,

sinθ = 12/13

cosθ = 5/13

Now , substituting

$\longrightarrow \sf\dfrac{2sin\theta-3cos\theta}{4sin\theta-9cos\theta}⟶ </p><p>$

$\displaystyle\sf\longrightarrow \dfrac{2\bigg(\dfrac{12}{13}\bigg)-3\bigg(\dfrac{5}{13}\bigg)}{4\bigg(\dfrac{12}{13}\bigg)-9\bigg(\dfrac{5}{13}\bigg)}⟶ </p><p>$

$\displaystyle\sf\longrightarrow \dfrac{\bigg(\dfrac{24}{13}\bigg)-\bigg(\dfrac{15}{13}\bigg)}{\bigg(\dfrac{48}{13}\bigg)-\bigg(\dfrac{45}{13}\bigg)}⟶ </p><p>$

$\displaystyle\sf\longrightarrow \dfrac{\bigg(\dfrac{24 - 15}{13}\bigg)}{\bigg(\dfrac{48 - 45}{13}\bigg)}⟶ </p><p>$

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

$\displaystyle\sf\longrightarrow \underline{\boxed{\textbf{\textsf{3}}}}⟶ </p><p></p><p> </p><p> </p><p>$

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