Question

Choose the correct answer from the alternatives given :
The value of $\frac{cos({90}^{\circ }+\theta )sec(-\theta )tan({180}^{\circ }-\theta )}{sec\theta sin({180}^{\circ }+\theta )cot(90-\theta )}$ is equal to

• A. 1
• B. $\pm 1$
• C. -1
• D. None of these

Answer 1

The correct option is C -1

$\frac{cos(90+\theta )sec(-\theta )tan(180-\theta )}{sec\theta sin(180+\theta )cot(90-\theta )}$
$=\frac{-sin\theta \times \frac{1}{cos\theta }\times \frac{sin(180-\theta )}{cos(180-\theta )}}{\frac{1}{cos\theta }\times (-sin\theta \times \frac{cos(90-\theta )}{sin(90-\theta )})}$
$=\frac{-sin\theta \times \frac{1}{cos\theta }\times \frac{sin\theta }{-cos\theta }}{\frac{1}{cos\theta }\times (-sin\theta \times \frac{sin\theta }{cos\theta })}=-1$