Question

If asec theta - btan theta=2 then find the value of bsec theta - atan theta=?

Answer 1

The value of $b\sec \theta - a\tan \theta$ = ± $\sqrt{a^2 -b^2-4}$

Step-by-step explanation:

We have,

$a\sec \theta - b\tan \theta=2$                       ...... (1)

Let $b\sec \theta - a\tan \theta=x$

To find, the value of $b\sec \theta - a\tan \theta=?$

Squaring equations (1) and (2) and subtracting them, we get

$a^2\sec^2 \theta + b^2\tan^2 \theta-2ab\sec \theta \tan \theta-(b^2\sec^2 \theta + a^2\tan^2 \theta-2ab\sec \theta \tan \theta)=2^2+x^2$

⇒ $a^2\sec^2 \theta + b^2\tan^2 \theta-2ab\sec \theta \tan \theta-b^2\sec^2 \theta -a^2\tan^2 \theta+2ab\sec \theta \tan \theta=4+x^2$

⇒ $a^2\sec^2 \theta + b^2\tan^2 \theta-b^2\sec^2 \theta -a^2\tan^2 \theta=4+x^2$

⇒ $a^2(\sec^2 \theta-\tan^2\theta) + b^2(\tan^2 \theta-\sec^2 \theta)=4+x^2$

Using the trigonometric identity,

$\sec^2 \theta-\tan^2\theta=1$

⇒ $a^2(1 + b^2(-1)=4+x^2$

⇒ $x^2=a^2 -b^2-4$

⇒ x = ± $\sqrt{a^2 -b^2-4}$

Thus, the value of $b\sec \theta - a\tan \theta$ = ± $\sqrt{a^2 -b^2-4}$