Question

evaluate: sinA cosA - sinA cos(90-A) cos A/ sec(90-A) - cosA sin(90-A) sinA/cosec(90-A)

Answer 1

please write question again .

Answer 2

Answer:

$\sin A \cos A - \frac{\sin A \cos(90-A)\cos A}{\sec(90-A)} - \frac{\cosA sin(90-A) \sin A}{\cosec(90-A)}=0$

Step-by-step explanation:

Given : $\sin A \cos A - \frac{\sin A \cos(90-A)\cos A}{\sec(90-A)} - \frac{\cosA sin(90-A) \sin A}{\csc(90-A)}$

To evaluate : The given expression?

Solution :

$\sin A \cos A - \frac{\sin A \cos(90-A)\cos A}{\sec(90-A)} - \frac{\cosA \sin(90-A) \sin A}{\csc(90-A)}$

$=\sin A \cos A - \frac{\sin A \sin A \cos A}{\csc A} - \frac{\cosA \cosA \sin A}{\sec A}$

[∵ sin(90°-a)=cosa, cos(90°-a)=sina , sec(90°-a) = coseca , cosec(90°-a) = seca]

$=\sin A \cos A -\sin^3 A \cos A-\cos^3A \sin A$

[∵1/seca = cosa , 1/coseca = sina]

$=\sin A \cos A -\sin A \cos A[\sin^2 A+\cos^2 A]$

$=\sin A \cos A -\sin A \cos A[1]$

[ ∵sin²Ф + cos²Ф = 1 ]

$=\sin A \cos A -\sin A \cos A$

$=0$