Question

sin theta = a/b then find sec theta + tan theta

Answer 1

Trigonometry

Let 'e' insted of theta ok
We have sine = a/b

Now,
sece + tane
= (1/cose)+(sine/cose)
= (1+sine)/cose
= (1+sine)/√(cos²e)
= (1+sine)/{√(1 - sin²e)}
= (1+a/b)/{√(1-a²/b²)
= (a+b)/√(b²-a²)

That's it
Hope it helped
=_=

Answer 2

Answer:

$sec\,\theta+tan\,\theta=\sqrt{\frac{b+a}{b-a}}$

Step-by-step explanation:

Given:

$sin\,\theta=\frac{a}{b}$

To find: $sec\,\theta+tan\,\theta$

Using Trigonometric identity,

$cos\,\theta=\sqrt{1-sin^2\,\theta}$

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

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

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

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

$\implies tan\,\theta=\frac{sin\,\theta}{cos\,\theta}=\frac{a}{\sqrt{b^2-a^2}}$

Now,

$sec\,\theta+tan\,\theta$

$=\frac{b}{\sqrt{b^2-a^2}}+\frac{a}{\sqrt{b^2-a^2}}$

$=\frac{b+a}{\sqrt{(b-a)(b+a)}}$

$=\frac{\sqrt{b+a}\sqrt{b+a}}{\sqrt{b-a}\sqrt{b+a}}$

$=\frac{\sqrt{b+a}}{\sqrt{b-a}}$

$=\sqrt{\frac{b+a}{b-a}}$

Therefore, $sec\,\theta+tan\,\theta=\sqrt{\frac{b+a}{b-a}}$