Question

if asin^2theta+bcos^2thetha=c ,then tan^2thetha=​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$tan^2 \theta =\frac{c-b}{a-c}$

Step-by-step explanation:

We know,

$tan^2\theta =\frac{asin^2\theta}{bcos^2\theta}$

Now from the given equation,

we can write,

$asin^2\theta +b(1-sin^2\theta)=c$

⇒$asin^2\theta +b-bsin^2\theta=c$

⇒$asin^2\theta-bsin^2\theta=c-b$

⇒$sin^2\theta(a-b)=c-b$

⇒$sin^2\theta=\frac{c-b}{a-b}$ $\\.....(i)$

Now again from the given equation,

we can write,

$a(1-cos^2\theta) +bcos^2\theta=c$

⇒$a-acos^2\theta +bcos^2\theta=c$

⇒$cos^2\theta(b-a)=c-a$

⇒$-cos^2\theta(a-b)=-(a-c)$

⇒$cos^2\theta(a-b)=a-c$

⇒$cos^2\theta=\frac{a-c}{a-b}$$\\ .......(ii)$

Now,

by dividing (i) by (ii) we get,

$\frac{sin^2\theta}{cos^2\theta}= (\frac{c-b}{a-b})(\frac{a-b}{a-c})$

⇒$tan^2\theta =\frac{c-b}{a-c}$

therefore, $tan^2\theta =\frac{c-b}{a-c}$ (Answer.)

hope this will help you....

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