Question

Which of the following do/does not reduce to unity?

• A. $\frac{\sin ({180}^0+A)}{\tan ({180}^0+A)}\frac{\cot ({90}^0+A)}{\tan ({90}^0+A)}\frac{\cos ({360}^0-A)\text{cosec}A}{\sin (-A)}$
• B. $\frac{\sin (-A)}{\sin ({180}^0+A)}-\frac{\tan ({90}^0+A)}{\cot A}+\frac{\cos A}{\sin ({90}^0+A)}$
• C. $\frac{\sin {24}^0\cos 6^0-\sin 6^0\cos {24}^0}{\sin {51}^0\cos {69}^0-\cos {51}^0\sin {69}^0}$
• D. $\frac{\cos ({90}^0+A)\sec (-A)\tan ({180}^0-A)}{\sec ({360}^0+A)\sin ({180}^0+A)\cot ({90}^0-A)}$

Answer 1

Answer: (B), (D)

$\frac{\sin (-A)}{\sin ({180}^0+A)}-\frac{\tan ({90}^0+A)}{\cot A}+\frac{\cos A}{\sin ({90}^0+A)}$
$\frac{\cos ({90}^0+A)\sec (-A)\tan ({180}^0-A)}{\sec ({360}^0+A)\sin ({180}^0+A)\cot ({90}^0-A)}$

Option A,

$\frac{\sin ({180}^0+A)}{\tan ({180}^0+A)}\frac{\cot ({90}^0+A)}{\tan ({90}^0+A)}\frac{\cos ({360}^0-A)\text{cosec}A}{\sin (-A)}$

$=\frac{-\sin A}{\tan A}\frac{-\tan A}{-\cot A}\frac{\cos A\text{cosec}A}{-\sin A}$

$=\frac{-\sin A}{\tan A}\frac{-\tan A}{-\cot A}\frac{\cot A}{-\sin A}$

$=1$

Now, option B,

$\frac{\sin (-A)}{\sin ({180}^0+A)}-\frac{\tan ({90}^0+A)}{\cot A}+\frac{\cos A}{\sin ({90}^0+A)}$

$=\frac{-\sin A}{-\sin A}-\frac{-\cot A}{\cot A}+\frac{\cos A}{\cos A}$

$=1+2=3$

Option C,

$\frac{\sin {24}^0\cos 6^0-\sin 6^0\cos {24}^0}{\sin {21}^0\cos {39}^0-\cos {51}^0\sin {69}^0}$

$=\frac{\sin (24-6)}{\sin (90-{69}^0)\cos (90-{51}^0)-\cos {51}^0\sin {69}^0}$

$=\frac{\sin (24-6)}{\cos {69}^0\sin {51}^0-\cos {51}^0\sin {69}^0}$

$=\frac{\sin 18}{\sin (69-51)}$

$=\frac{\sin 18}{\sin 18}$

$=1$

Option D

$\frac{\cos ({90}^0+A)\sec (-A)\tan ({180}^0-A)}{\sec ({360}^0+A)\sin ({180}^0+A)\cot ({90}^0-A)}$

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

$=-1$

Option B and D does not reduce to $1$