Question

PLZ SOLVE

If a²sec²∅ - b²tan²∅ = c² , prove that

sin²∅ = c² - a²/c² - b²


IF DON'T KNOW THEN DON'T ANSWER

Answer 1

$\bold{TRIGONOMETRY\:FACTS :-}$

✔ Trigonometry is the study of relationships between the sides and the angles of a triangle.

✔ The general trigonometric ratios are sin theta, cos theta, tan theta, sec theta, cosec theta and cot theta.

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We all know a triangle has three parts : base, height and hypotenuse.

✔ Sin theta = height/hypotenuse.

✔ Cos theta = base/hypotenuse.

✔ tan theta = height /base.

✔ cot theta = base/height.

✔ sec theta = hypotenuse/base.

✔ cosec theta = hypotenuse/height.

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$\bold{\huge{SOLUTION \: :}}$

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a²sec² θ-b²tan² θ = c²

Or, a²sec²θ-b²tan²θ = c × 1

Or, a²sec²θ-b²tan²θ = c²sec²θ-c²tan²θ

Or, sec²θ(a²-c²) - tan²θ (b²-c²)

Or, $\frac{ {sec}^{2}θ}{ {tan}^{2}θ} = \frac{ {b}^{2} - {c}^{2} }{ {a}^{2} - {c}^{2} }$

Or, $\frac{1}{ {sin}^{2}θ} = \frac{ {b}^{2} - {c}^{2} }{ {a}^{2} - {c}^{2} }$

Or, ${sin}^{2}θ \: = \: \frac{ - ( - {a}^{2} + {c}^{2}) }{ - ( - {b}^{2} + {c}^{2} )}$

Or, ${sin}^{2}θ \: = \frac{ {c}^{2} - {a}^{2} }{ {c}^{2} - {b}^{2} }$

THUS, IT CAN BE PROVED.