Question

10. If sin 0 = then prove that (sec @ + tan 0) =
V 6-a
bta
b​

Answer 1

Answer:

Given:

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

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

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,

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

$</p><p>=\frac{b}{\sqrt{b^2-a^2}}+\frac{a}{\sqrt{b^2-a^2}}$

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

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

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

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

Hence Proved