Question

Prove that $\frac{cos(\pi - A) cot[\frac{\pi}{2}+ A] cos(-A)}{tan(\pi + A) tan[\frac{3\pi}{2}+ A] sin(2\pi - A)}$ = cos A.

Answer 1

we know, from trigonometric concepts

cos(π - A) = -cosA

cos(-A) = cosA

cot(π/2 + A) = -tanA

tan(π + A) = tanA

tan(3π/2 + A) = -cotA

sin(2π - A) = -sinA

now, LHS = {cos(π- A)cot(π/2 + A) cos(-A)}/{tan(π + A) tan(3π/2 + A) sin(2π - A)}

= {(-cosA)(-tanA)cosA}/{tanA(cotA)(-sinA)}

= {-cos²A}/{-cotAsinA}

= {cos²A}/{cosA/sinA × sinA}

= cosA = RHS

Answer 2

HELLO DEAR,



WE KNOW:-cos(π - A) = -cosA
cos(-A) = cosA , cot(π/2 + A) = -tanA
tan(π + A) = tanA
tan(3π/2 + A) = -cotA
sin(2π - A) = -sinA

now,
{cos(π- A)cot(π/2 + A) cos(-A)}/{tan(π + A) tan(3π/2 + A) sin(2π - A)}

=> {(-cosA)(-tanA)cosA}/{tanA(cotA)(-sinA)}

=> {-cos²A}/{-cotAsinA}

=> {cos²A}/{cosA/sinA * sinA}

=> cosA


I HOPE IT'S HELP YOU DEAR,
THANKS