Question

If sec theta=13/12 then find the value of root 1-sin^2 theta​

Answer 1

Answer:

secx=13/12

cosx= 12/13

sinx=5/13

we want , ......where x=theta

(1-sin^2x)^2=1-25/169

=169-25/169

=144/169

=√144/√169

=12/13

=cosx=cos theta

Answer 2

The value of $\sqrt{1-\sin^2 \theta}$ = $\dfrac{12}{13}$

Step-by-step explanation:

We have,

$\sec \theta=\dfrac{13}{12}$     ..... (1)

To find, the value of $\sqrt{1-\sin^2 \theta}$ = ?

∴ $\sqrt{1-\sin^2 \theta}$

Using the trigonometric identity,

$\sin^2 A+\cos^2 A$ = 1

⇒ $\cos^2 A= 1 -\sin^2 A$

= $\sqrt{\cos^2 \theta}$

= $\cos \theta$

Using the trigonometric identity,

$\cos A=\dfrac{1}{\sec A}$

Using equation (1), we get

= $\dfrac{12}{13}$

∴ The value of $\sqrt{1-\sin^2 \theta}$ = $\dfrac{12}{13}$

Thus, the value of $\sqrt{1-\sin^2 \theta}$ = $\dfrac{12}{13}$