Question

If sin=a^2-b^2/a^2+b^2 then find the other five trigonometric ratios​

Answer 1

As We know that sin ∅ = a² - b²/a² + b² .

$\bigstar \ {\underline{\sf{ Concept }}}$

The most important task of trigonometry to find the remaining sides and angles of a triangle when some of its sides and angles are given.

$\circ \ {\underline{\boxed{\sf\orange{ Sin \ \theta = \dfrac{Perpendicular}{Hypotenuse} }}}}$

According to Question, We have that:

$\colon\implies{\sf{ sin \ \theta = \dfrac{a^2-b^2}{a^2+b^2} }}$

So, Draw a right Triangle & Right angled at B as such that.

• Perpendicular = a² - b²

• Hypotenuse = a² + b²

• ${\sf{ \angle BAC = \theta }}$

By Using Pythagoras theorem to Find the desired results as:

$\colon\implies{\tt{ AC^2 = AB^2 + BC^2 }} \\ \\ \\ \colon\implies{\tt{ (a^2 + b^2 )^2 = AB^2 + (a^2 - b^2 )^2 }} \\ \\ \\ \colon\implies{\tt{ AB^2 = (a^2 + b^2 )^2 - (a^2 - b^2 )^2 }} \\ \\ \\ \colon\implies{\tt{ AB^2 = (a^4 + b^4 + 2a^2b^2 ) - (a^4 + b^4 - 2a^2b^2 ) }} \\ \\ \\ \colon\implies{\tt{ AB^2 = 4a^2b^2 = (2ab)^2 }} \\ \\ \\ \colon\implies{\tt{ AB = 2ab }}$

When we consider the trigonometric ratios of ${\sf{ \angle BAC = \theta }}$ , we have:

• Base = AB = 2ab

• Perpendicular = BC = a² - b²

• Hypotenuse = AC = a² + b²

As we also know that we have to Find the other Five Trigonometric ratio as:

$\circ \ {\underline{\boxed{\sf{ cos \ \theta = \dfrac{Base}{Hypotenuse} = \dfrac{2ab}{a^2+b^2} }}}}$

$\circ \ {\underline{\boxed{\sf{ tan \ \theta = \dfrac{Perpendicular}{Base} = \dfrac{a^2-b^2}{2ab} }}}}$

$\circ \ {\underline{\boxed{\sf{cosec \ \theta = \dfrac{Hypotenuse}{Perpendicular} = \dfrac{a^2+b^2}{a^2-b^2} }}}}$

$\circ \ {\underline{\boxed{\sf{ sec \ \theta = \dfrac{Hypotenuse}{Base} = \dfrac{a^2+b^2}{2ab} }}}}$

$\circ \ {\underline{\boxed{\sf{ cot \ \theta = \dfrac{Base}{Perpendicular} = \dfrac{2ab}{a^2-b^2} }}}}$

• It should be noted that sin ∅ is an abbreviation for " sine of angle ∅ ", It is not the Product of sin and ∅.