Question

if sin theta equals to 12/13 then find the value of sin squared theta minus cos squared theta upon 2 sin squared theta into cos theta into 1 upon ten squared theta

Answer 1

Here is your answer

Answer 2

$\dfrac{\sin^2 \theta-\cos^2 \theta}{2\sin \theta\cos \theta} \times \dfrac{1}{\tan^2 \theta}=\dfrac{119}{288}$

Step-by-step explanation:

We have,

$\sin \theta=\dfrac{12}{13}$

To find, $\dfrac{\sin^2 \theta-\cos^2 \theta}{2\sin \theta\cos \theta} \times \dfrac{1}{\tan^2 \theta}=?$

∴ $\sin \theta=\dfrac{12}{13}=\dfrac{p}{h}$

∴$b=\sqrt{13^{2}-12^{2}} =\sqrt{169-144}=5$

Where, p = perpendicular, b = base and h = hypotaneous

=$\dfrac{(\dfrac{12}{13} )^{2} -(\dfrac{5}{13} )^{2} }{2(\dfrac{12}{13} (\dfrac{5}{13} )} \times \dfrac{1}{\dfrac{12}{5}}$

$=\dfrac{\dfrac{144}{169}  -\dfrac{25}{169}}{\dfrac{120}{169} } \times \dfrac{1}{\dfrac{12}{5}}$

$=\dfrac{\dfrac{144-25}{169}}{\dfrac{120}{169} } \times \dfrac{1}{\dfrac{12}{5}}$

$=\dfrac{119}{120} \times \dfrac{1}{\dfrac{12}{5}}$

$=\dfrac{119}{288}$

Henec, $\dfrac{\sin^2 \theta-\cos^2 \theta}{2\sin \theta\cos \theta} \times \dfrac{1}{\tan^2 \theta}=\dfrac{119}{288}$