Question

Which of the following options is/are true?
$[A\in (0^{\circ },{90}^{\circ })]$

• A. $\frac{\sin ({180}^{\circ }+A)}{\tan ({180}^{\circ }+A)}\frac{\cot ({90}^{\circ }+A)}{\tan ({90}^{\circ }+A)}\frac{\cos ({360}^{\circ }-A)\text{cosec}A}{\sin (-A)}=1$
• B. $\frac{\sin (-A)}{\sin ({180}^{\circ }+A)}-\frac{\tan ({90}^{\circ }+A)}{\cot A}+\frac{\cos A}{\sin ({90}^{\circ }+A)}=1$
• C. $\frac{\sin {24}^{\circ }\cos 6^{\circ }-\sin 6^{\circ }\cos {24}^{\circ }}{\sin {21}^{\circ }\cos {39}^{\circ }-\cos {51}^{\circ }\sin {69}^{\circ }}=-1$
• D. $\frac{\cos ({90}^{\circ }+A)\sec (-A)\tan ({180}^{\circ }-A)}{\sec ({360}^{\circ }+A)\sin ({180}^{\circ }+A)\cot ({90}^{\circ }-A)}=-1$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $\frac{\sin ({180}^{\circ }+A)}{\tan ({180}^{\circ }+A)}\frac{\cot ({90}^{\circ }+A)}{\tan ({90}^{\circ }+A)}\frac{\cos ({360}^{\circ }-A)\text{cosec}A}{\sin (-A)}=1$
C $\frac{\sin {24}^{\circ }\cos 6^{\circ }-\sin 6^{\circ }\cos {24}^{\circ }}{\sin {21}^{\circ }\cos {39}^{\circ }-\cos {51}^{\circ }\sin {69}^{\circ }}=-1$
D $\frac{\cos ({90}^{\circ }+A)\sec (-A)\tan ({180}^{\circ }-A)}{\sec ({360}^{\circ }+A)\sin ({180}^{\circ }+A)\cot ({90}^{\circ }-A)}=-1$

$\left(a\right)$
$\frac{\sin ({180}^{\circ }+A)}{\tan ({180}^{\circ }+A)}\frac{\cot ({90}^{\circ }+A)}{\tan ({90}^{\circ }+A)}\frac{\cos ({360}^{\circ }-A)\text{cosec}A}{\sin (-A)}$
$=\frac{-\sin A}{\tan A}\frac{(-\tan A)}{(-\cot A)}\frac{\cos A\text{cosec}A}{(-\sin A)}$
$=1$

$\left(b\right)$
$\frac{\sin (-A)}{\sin ({180}^{\circ }+A)}-\frac{\tan ({90}^{\circ }+A)}{\cot A}+\frac{\cos A}{\sin ({90}^{\circ }+A)}$
$=\frac{-\sin A}{-\sin A}-\frac{-\cot A}{\cot A}+\frac{\cos A}{\cos A}$
$=3$

$\left(c\right)$
$\frac{\sin {24}^{\circ }\cos 6^{\circ }-\sin 6^{\circ }\cos {24}^{\circ }}{\sin {21}^{\circ }\cos {39}^{\circ }-\cos {51}^{\circ }\sin {69}^{\circ }}$
$=\frac{\sin {24}^{\circ }\cos 6^{\circ }-\sin 6^{\circ }\cos {24}^{\circ }}{\sin {21}^{\circ }\cos {39}^{\circ }-\sin {39}^{\circ }\cos {21}^{\circ }}$
$=\frac{\sin ({24}^{\circ }-6^{\circ })}{\sin ({21}^{\circ }-{39}^{\circ })}=\frac{\sin 18°}{-\sin 18°}=-1$

$\left(d\right)$
$\frac{\cos ({90}^{\circ }+A)\sec (-A)\tan ({180}^{\circ }-A)}{\sec ({360}^{\circ }+A)\sin ({180}^{\circ }+A)\cot ({90}^{\circ }-A)}$
$=\frac{-\sin A\sec A(-\tan A)}{\sec A(-\sin A)\tan A}$
$=-1$