Question

If $sec\Theta =\frac{13}{5}$, show that $\frac{2sin\Theta-3cos\Theta}{4sin\Theta-9cos\Theta}=3$.

Answer 1

SOLUTION IS IN THE ATTACHMENT.

** Trigonometry is the study of the relationship between the sides and angles of a triangle.

The ratio of the sides of a right angled triangle with respect to its acute angles are called trigonometric ratios.

** For any acute angle in a right angle triangle the side opposite to the acute angle is called a perpendicular(P),  the side adjacent to this acute angle is called the base(B) and side opposite to the right angle is called the hypotenuse(H).

** Find the third  side of the right ∆ ABC by using Pythagoras theorem (AC² = AB² + BC²).

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Hello there !

Due to inconvenience i am writing theta as A.

Given thing, sec A = 13 / 5

We know that sec A is actually the ratio of hypotenuse to base( for angle A ). Mathematically we can say that secA = $\dfrac{13}{5}$

Now,

secA = $\dfrac{13}{5}$

$\dfrac{hypotenuse}{base for angle A } = \dfrac{13}{5}$

Now, let hypotenuse of the triangle be 13x and base of the triangle be 5x.

As it is a trigonometry of right angled triangles, we can apply pythagoras theorem, this theorem say that the square of length of hypotenuse is equal to the sum of the squares of the remaining sides of the triangle.

                   By Pythagoras theorem

( 5x )^2 + ( height )^2 = ( 13 x )^2

hieght^2 = 169x^2 - 25x^2

height^2 = 144x^2

height = $\pm 12 x$

height of any triangle cannot be negative, so height = 12 x

So,

sinA = height / hypotenuse [ from trigonometric table ]

sinA = 12 x / 13 x

sinA = 12 / 13

cosA = base / hypotenuse [ from  trigonometric table ]

cosA = 1 / secA

cosA = 5 / 13

Therefore,

$\dfrac{2sin\Theta-3cos\Theta}{4sin\Theta-9cos\Theta}$

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

⇒ $\dfrac{\dfrac{24}{13} - \dfrac{15}{13}}{\dfrac{48}{13} - \dfrac{45}{13}}$

⇒ $\dfrac{\bigg( \dfrac{1}{13}\bigg) \bigg( 24 - 15\bigg)}{\bigg( \dfrac{1}{13} \bigg) (48-45)}$

⇒ $\dfrac{9}{3}$

⇒ ${3}$

There are few more methods for solving, if you want comment.