Question

The value of $\frac{2(sin2\theta +2co{s}^2\theta -1)}{cos\theta -sin\theta -cos3\theta +sin3\theta }$ is

• A. $cos\theta$
• B. $sec\theta$
• C. $cosec\theta$
• D. $sin\theta$

Answer 1

The correct option is C

$cosec\theta$

We have,

$\frac{2(sin2\theta +2co{s}^2\theta -1)}{cos\theta -sin\theta -cos3\theta +sin3\theta }=\frac{2(sin2\theta +cos2\theta )}{cos\theta -sin\theta -4co{s}^3\theta +3cos\theta +3sin\theta -4si{n}^3\theta }=\frac{2(sin2\theta +cos2\theta )}{4cos\theta -4co{s}^3\theta +2sin\theta -4si{n}^3\theta }=\frac{2(sin2\theta +cos2\theta )}{4cos\theta (1-co{s}^2\theta )+2sin\theta (1-2si{n}^2\theta )}=\frac{2(sin2\theta +cos2\theta )}{4cos\theta si{n}^2\theta +2sin\theta cos2\theta }$

$=\frac{2(sin2\theta +cos2\theta )}{2\times 2sin\theta cos\theta sin\theta +2sin\theta cos2\theta }=\frac{2(sin2\theta +cos2\theta )}{2sin2\theta sin\theta +2sin\theta cos2\theta }=\frac{2(sin2\theta +cos2\theta )}{2sin2\theta (sin2\theta +cos2\theta )}=\frac{1}{sin\theta }=cosec\theta$