Question

sin theta is equal to a by b find sec theta + tan theta in terms of a and b ​

Answer 1

Answer:

Step-by-step explanation:

Given

Sin theta=a÷b

Take a right angle triangle

Use pythogerous therom

b2-a2=x2

X=root b2-a2

Sectheta= hyp÷adj

Sectheta=b÷root b2-a2

Tantheta=opp÷adj

tantheta=a÷root b2-a2

Sectheta+tantheta=a+b÷root b2-a2

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

We know that,

sin² x + cos² x = 1

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

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

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

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

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

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

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