Question

prove that sin (180+A)*cos(90-A)*tan(270-A)/sec(540-A)*cos(360+A)*cosec(270+A)=-sin A cos^2 A

Answer 1

sin^3(180-A) tan(360-A) sec^2(180-A)/cos^2(90+A) cos^2A sin(180-A)=tan^3A

Answer 2

$\dfrac{sin (180+A)cos(90-A)tan(270-A)}{sec(540-A)cos(360+A)cosec(270+A)}=-sin A cos^2 A$ is proved.

Step-by-step explanation:

To Prove: $\dfrac{sin (180+A)cos(90-A)tan(270-A)}{sec(540-A)cos(360+A)cosec(270+A)}=-sin A cos^2 A$

Solution:

• Proof of the given trigonometric equation

Considering LHS such that we have

$\Rightarrow \dfrac{sin (180+A)cos(90-A)tan(270-A)}{sec(540-A)cos(360+A)cosec(270+A)}$

$\Rightarrow \dfrac{-sin (A)cos(90-A)tan(270-A)}{sec(540-A)cos(360+A)cosec(270+A)} \ \ \because sin(180+A) = -sin(A)$

$\Rightarrow \dfrac{-sin (A)sin(A)tan(270-A)}{sec(540-A)cos(360+A)cosec(270+A)} \ \ \because cos(90-A) = sin(A)$

$\Rightarrow \dfrac{-sin (A)sin(A)cot(A)}{sec(540-A)cos(360+A)cosec(270+A)} \ \ \because tan(270-A) = cot(A)$

Since $sec(540-A) = sec(360 + 180-A) = sec (180-A) = -sec(A)$, we can write,

$\Rightarrow \dfrac{-sin (A)sin(A)cot(A)}{-sec(A)cos(360+A)cosec(270+A)}$

$\Rightarrow \dfrac{sin (A)sin(A)cot(A)}{sec(A)cos(A)cosec(270+A)} \ \ \because cos(360+A) = cos(A)$

$\Rightarrow \dfrac{sin (A)sin(A)cot(A)}{sec(A)cos(A)( -sec(A))} \ \ \because cosec(270+A) = -sec(A)$

$\Rightarrow - \dfrac{sin (A) \times sin(A) \times \dfrac{cos(A)}{sin(A)} }{\dfrac{1}{cos(A)} \times cos(A) \times \dfrac{1}{cos(A)}}$

Solving the above, you will get

$\Rightarrow - sinAcos^2A = RHS$

Hence, proved $\dfrac{sin (180+A)cos(90-A)tan(270-A)}{sec(540-A)cos(360+A)cosec(270+A)}=-sin A cos^2 A$