Question

if sin theta =12/13,find the value of sin^2theta-cos^2theta/2sin theta cos theta

Answer 1

This question answer

Answer 2

Answer:

$\frac{\sin^2\theta-\cos^2\theta}{2\sin \theta cos \theta}=\frac{119}{120}$

Step-by-step explanation:

Given : If $\sin\theta =\frac{12}{13}$

To find : The value of $\frac{\sin^2\theta-\cos^2\theta}{2\sin \theta cos \theta}$

Solution :

$\frac{\sin^2\theta-\cos^2\theta}{2\sin \theta cos \theta}$

We know, $\sin\theta =\frac{12}{13}=\frac{P}{H}$

We can find the base by Pythagoras theorem,

$B=\sqrt{H^2-P^2}$

$B=\sqrt{13^2-12^2}$

$B=\sqrt{169-144}$

$B=\sqrt{25}$

$B=5$

So, $\cos\theta =\frac{B}{H}=\frac{5}{13}$

Substitute in the value,

$=\frac{(\frac{12}{13})^2-(\frac{5}{13})^2}{2(\frac{12}{13})(\frac{5}{13})}$

$=\frac{(12^2-5^2)\times (13\times13)}{(13\times 13)(2\times 12\times 5)}$

$=\frac{144-25}{2\times 12\times 5}$

$=\frac{119}{120}$

Therefore, $\frac{\sin^2\theta-\cos^2\theta}{2\sin \theta cos \theta}=\frac{119}{120}$