Question

If sin theta = 2ab/a²+b² then find out sec theta + tan theta

Answer 1

Answer:

here sin theta =p/h=2ab/a^2+b^2

so b =root (h^2-p^2)= root (a^4+b^4+2a^2b^2-4a^2b^2)=root(a^4+b^4-2a^2b^2)=(a^2-b^2)

so ,sec theta+tan theta = b/h+p/b=

(a^2-b^2)/(a^2+b^2)+2ab/(a^2-b^2)=[a^4+b^4-2a^2b^2+2ab(a^2+b^2)]/(a^4-b^4)=now solve it yourself u get ans.

Answer 2

Required value is $\dfrac{a-b}{a+b}$

Step-by-step explanation:

Since we have given that

$\sin \theta=\dfrac{2ab}{a^2+b^2}$

So, here , P = $2ab$

H = $a^2+b^2$

So, B would be

$\sqrt{H^2-P^2}\\\\=\sqrt{(a^2+b^2)^2-(2ab)^2}\\\\=\sqrt{a^4+b^4+2a^2b^2-4a^2b^2}\\\\=\sqrt{a^4+b^2-2a^2b^2}\\\\=\sqrt{(a^2-b^2)^2}=a^2-b^2$

So, we get that

$\sec \theta+\tan \theta\\\\=\dfrac{H}{B}+\dfrac{P}{B}\\\\=\dfrac{a^2+b^2}{a^2-b^2}-\dfrac{2ab}{a^2-b^2}\\\\=\dfrac{a^2+b^2-2ab}{a^2-b^2}\\\\=\dfrac{(a-b)^2}{(a-b)(a+b)}\\\\=\dfrac{a-b}{a+b}$

Hence, required value is $\dfrac{a-b}{a+b}$

# learn more:

If sin theta is equal to 2ab by a square plus b square.

Find the value of

(a) cos theta

(b) tan theta

(c) cosec theta

(d) sec theta

(e) cot theta

https://brainly.in/question/10620154