Question

If $tan\theta -cot\theta =2\sqrt{3}$ (where ${45}^{\circ }<\theta <{90}^{\circ }$, then $\frac{si{n}^3\theta -co{s}^3\theta }{sin\theta cos\theta (sin\theta -cos\theta )}$ equals

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is D 5

$\frac{si{n}^3\theta -co{s}^3\theta }{sin\theta cos\theta (sin\theta -cos\theta )}$

=$\frac{(sin\theta -cos\theta )(si{n}^2\theta +co{s}^2\theta +sin\theta cos\theta )}{sin\theta cos\theta (sin\theta -cos\theta )}$

$=(tan\theta +cot\theta +1)$ ......(i)

Now , $(tan\theta +cot\theta )=\sqrt{(tan\theta -cot\theta {)}^2+4}$ [$(a+b{)}^2=(a-b{)}^2+4ab$]

$=\sqrt{(2\sqrt{3}{)}^2+4}$

=4 .....(ii)

Putting this value in eqn. (i), we get

$\frac{si{n}^3\theta -co{s}^3\theta }{sin\theta cos\theta (sin\theta -cos\theta )}$ =4+1

=5