Question

$\frac{\cos ({90}^{\circ }+\theta )\sec (-\theta )\tan ({180}^{\circ }-\theta )}{\sin ({360}^{\circ }+\theta )\sec ({180}^{\circ }+\theta )\cot ({90}^{\circ }-\theta )}$ is equal to

• A. $2$
• B. $1$
• C. $-1$
• D. $0$

Answer 1

Answer: (C)

$-1$

Formulae:

$\cos (90-\theta )=\sin \theta$

$\sin (90-\theta )=\cos \theta$

$\tan (90-\theta )=\cot \theta$

$\cot (90-\theta )=\tan \theta$

$\sec (90-\theta )=\csc \theta$

$cosec(90-\theta )=\sec \theta$

$\frac{\cos (90°+\theta )\sec (-\theta )\tan (180°-\theta )}{\sin (360°+\theta )\sec (180°+\theta )\cot (90°-\theta )}$

$=\frac{-\sin \theta .\sec \theta (-\tan \theta )}{\sin \theta .(-\sec \theta )\tan \theta }$

$=-1.$

Hence, the answer is $-1.$