Question

Evaluate:$\frac{2sin\theta cos\theta -cos\theta }{1-sin\theta +{sin}^2\theta -{cos}^2\theta }$

• A. $\cos \theta$
• B. $1$
• C. $\tan \theta$
• D. $\cot \theta$

Answer 1

Answer: (C)

$\tan \theta$

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

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

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

$=\frac{cos\theta (2sin\theta -1)}{sin\theta (2sin\theta -1)}$

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