Question

If $a^2{\sec }^2\theta -b^2{\tan }^2\theta =c^2.$ Find the value of $\sin \theta$

• A. $\pm \sqrt{\frac{c^2-b^2}{c^2-a^2}}$
• B. $\pm \sqrt{\frac{c^2+2b^2}{c^2-2a^2}}$
• C. $\pm \sqrt{\frac{c+b}{c^2+a^2}}$
• D. $\pm \sqrt{\frac{c^2-a^2}{c^2-b^2}}$

Answer 1

The correct option is D $\pm \sqrt{\frac{c^2-a^2}{c^2-b^2}}$

Given,

$a^2{\sec }^2\theta -b^2{\tan }^2\theta =c^2.$

or, $\frac{a^2-b^2{\sin }^2\theta }{{\cos }^2\theta }=c^2$

or, $a^2-b^2{\sin }^2\theta =c^2(1-{\sin }^2\theta )$ ........ [ Since ${\cos }^2\theta =1-{\sin }^2\theta$]

or, $a^2-c^2=(b^2-c^2){\sin }^2\theta$

or, $\sin \theta =\pm \sqrt{\frac{c^2-a^2}{c^2-b^2}}$