Question

Find the value of $\frac{sin\theta -2si{n}^3\theta }{2co{s}^3\theta -cos\theta }$.

• A. $2cot\theta$
• B. $sin\theta$
• C. $\frac{tan\theta }{2}$
• D. $tan\theta$

Answer 1

The correct option is D $tan\theta$

Given: $\frac{sin\theta -2si{n}^3\theta }{2co{s}^3\theta -cos\theta }$

$=\frac{sin\theta (1-2si{n}^2\theta )}{cos\theta (2co{s}^2\theta -1)}$

We know that, $si{n}^2\theta +co{s}^2\theta =1$ and $\frac{sin\theta }{cos\theta }=tan\theta$.

$=\frac{tan\theta (si{n}^2\theta +co{s}^2\theta -2si{n}^2\theta )}{(2co{s}^2\theta -(si{n}^2\theta +co{s}^2\theta \left)\right)}$

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

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

Cancelling out the common terms, we get,

$=tan\theta$