Question

Answer by formula please ​

Answer 1

Answer:

Step-by-step explanation:

I honestly have no idea what you mean by answer by formula, but I'm going to give it my best. I began by squaring both sides to get:

(a² - b²) tan²θ = b² and then distributed to get:

a² tan²θ - b² tan²θ = b² and then got the b terms on the side to get:

a² tan²θ = b² + b² tan²θ and then changed the tans to sin/cos to get:

$\frac{a^2sin^2\theta}{cos^2\theta}=b^2+\frac{b^2sin^2\theta}{cos^2\theta}$ and isolated the sin-squared on the left to get:

$a^2sin^2\theta=cos^2\theta(b^2+\frac{b^2sin^2\theta}{cos^2\theta})$ and distributed to get:

***$a^2sin^2\theta=b^2cos^2\theta+b^2sin^2\theta$*** and factored the right side to get:

$a^2sin^2\theta=b^2(sin^2\theta+cos^2\theta)$ and utilized a trig Pythagorean identity to get:

$a^2sin^2\theta=b^2(1)$ and then solved for sinθ in the following way:

$sin^2\theta=\frac{b^2}{a^2}$ so

$sin\theta=\frac{b}{a}$ This, along with the *** expression above will be important. I'm picking up at the *** to solve for cosθ:

$a^2sin^2\theta=b^2cos^2\theta+b^2sin^2\theta$ and get the cos²θ alone on the right by subtracting to get:

$a^2sin^2\theta-b^2sin^2\theta=b^2cos^2\theta$ and divide by b² to get:

$\frac{a^2sin^2\theta}{b^2}-sin^2\theta=cos^2\theta$ and factor on the left to get:

$sin^2\theta(\frac{a^2}{b^2}-1)=cos^2\theta$ and take the square root of both sides to get:

$\sqrt{sin^2\theta(\frac{a^2}{b^2}-1) }=cos\theta$ and simplify to get:

$\frac{sin\theta}{b}\sqrt{a^2-b^2}=cos\theta$ and go back to the identity we found for sinθ and sub it in to get:

$\frac{\frac{b}{a} }{b}\sqrt{a^2-b^2}=cos\theta$ and simplifying a bit gives us:

$\frac{1}{a}\sqrt{a^2-b^2}=cos\theta$

That's my spin on things....not sure if it's what you were looking for. If not.....YIKES