Question

Simplify the expression $\dfrac{\cos (2\pi +A)\cdot \cos ec(2\pi +A)\cdot \tan (2\pi +A)}{\sec (2\pi +A).\sin A.\cot (\pi +A)}$
A. 0
B. 2
C. 1
D. 3

Answer 1

Hint: We can solve this question by using trigonometric periodicity identities.
These are $\cos (2\pi +A)=\cos A$ , $\cos ec(2\pi +A)=\cos ecA$,$\tan (2\pi +A)=\tan A$,$\sec (2\pi +A)=\sec A$, $\cot (2\pi +A)=\cot A$

Complete step by step solution:
Given expression is $\dfrac{\cos (2\pi +A)\cdot \cos ec(2\pi +A)\cdot \tan (2\pi +A)}{\sec (2\pi +A).\sin A.\cot (\pi +A)}$
By using trigonometric periodicity identities we can write given expression as
$\Rightarrow \dfrac{\cos (2\pi +A)\cdot \cos ec(2\pi +A)\cdot \tan (2\pi +A)}{\sec (2\pi +A).\sin A.\cot (\pi +A)}=\dfrac{\cos A\cdot \cos ecA\cdot \tan A}{\sec A\cdot \sin A\cdot \cot A}$
$\Rightarrow \dfrac{\cos (2\pi +A)\cdot \cos ec(2\pi +A)\cdot \tan (2\pi +A)}{\sec (2\pi +A).\sin A.\cot (\pi +A)}=\dfrac{\cos A\cdot \dfrac{1}{\sin A}\cdot \tan A}{\dfrac{1}{\cos A}\cdot \sin A\cdot \cot A}$ \left\\{ \because \cos ecA=\dfrac{1}{\sin A},\sec A=\dfrac{1}{\cos A} \right\\}

$\Rightarrow \dfrac{\cos (2\pi +A)\cdot \cos ec(2\pi +A)\cdot \tan (2\pi +A)}{\sec (2\pi +A).\sin A.\cot (\pi +A)}=\dfrac{\cot A\cdot \tan A}{\tan A\cdot \cot A}$ \left\\{ \because \dfrac{\cos A}{\sin A}=\cot A,\dfrac{\sin A}{\cos A}=\tan A \right\\}
$\Rightarrow \dfrac{\cos (2\pi +A)\cdot \cos ec(2\pi +A)\cdot \tan (2\pi +A)}{\sec (2\pi +A).\sin A.\cot (\pi +A)}=1$
Hence option C is correct.

Note:In general periodic trigonometric functions are that which repeats it’s value after certain values of angles.
Sin and Cos are periodic functions with period $2\pi$.
Period of tan is $\pi$.