Question

Sin theta = 12/13 . Find the vaue of

Sin^2 -cos^2/2 sin.cos × 1/ tan^2

Answer 1

We are given with,

✏$\bold{sin\theta = \frac{12}{13}}$

✏ We know that,  $\bold{sin\theta = \frac{Perpendicular}{Hypotenuse}}$

•°• Perpendicular = 12 x {Assume.}

•°• Hypotenuse = 13 x

•°• By Pythagoras Theorem, We have ;

$\bold{H^{2} = P^{2} + B^{2}}$

Where, H =  Hypotenuse . P = Perpendicular. B = Base of a right angled triangle.

By applying values, we get ;

$\bold{\sqrt169x^{2} - 144x^{2} = B}$

$\bold{\sqrt25x^{2} = B}$

•°•  B = Base = 5x

Hence, We get the value of $\bold{cos\theta\:\:as,\frac{5x}{13x}}$

•°• $\bold{cos\theta = \frac{5\cancel{x}}{13\cancel{x}}}$

•°• $\bold{cos\theta = \frac{5}{13}}$

Now, Putting Values in Question ,

✏$\bold{\frac{sin^{2}\theta\:-\:cos^{2}\theta}{2\times\:sin\theta\times\:cos\theta}\:\times\frac{1}{tan^{2}\theta}}$

✏$\bold{\frac{(\frac{12}{13})^{2}\:-\:(\frac{5}{13})^{2}}{2\times\frac{12}{13}\times\frac{5}{13}}\:\times\frac{1}{\frac{sin^{2}\theta}{cos^{2}\theta}}}$

✏$\bold{\frac{(\frac{12}{13})^{2}\:-\:(\frac{5}{13})^{2}}{2\times\frac{12}{13}\times\frac{5}{13}}\:\times\frac{cos^{2}\theta}{sin^{2}\theta}}}$

✏$\bold{\frac{(\frac{12}{13})^{2}\:-\:(\frac{5}{13})^{2}}{2\times\frac{12}{13}\times\frac{5}{13}}\:\times(\frac{5}{12})^{2}}$

✏$\bold{\frac{\frac{144-25}{169}}{2\times\frac{12}{13}\times\frac{5}{13}}\:\times\frac{25}{144}}$

✏$\bold{\frac{\frac{119}{169}}{\frac{24\times5}{169}}\:\times\frac{25}{144}}$

✏$\bold{\frac{\frac{119}{\cancel{169}}}{\frac{120}{\cancel{169}}}\:\times\frac{25}{144}}$

✏$\bold{\frac{\frac{119}{1}}{\frac{120}{1}}\:\times\frac{25}{144}}$

✏$\bold{\frac{119}{120}\:\times\frac{25}{144}}$

= $\bold{\frac{595}{3456}}$ (Multiplying Numerators and Denominators.)

Thanks :)