Question

Prove that $\frac{sin(3\pi - A) cos[A - \frac{\pi}{2}] tan(\frac{3\pi}{2} - A)}{cosec(\frac{13\pi}{2} + A) sec(3\pi + A) cot(A - \frac{\pi}{2})}$ = cos⁴ A

Answer 1

sin(3π - A) = sin{2π + ( π - A)}
= sin(π - A) = sinA

cos(A - π/2) = cos{-(π/2 - A)} = cos(π/2 - A)[as we know, cos(-x) = cosx]
= cos(π/2 - A) = sinA

tan(3π/2 - A) = cotA [ in 3rd quadrant, tan, cot are positive ]

cosec(13π/2 + A) = cosec(6π + π/2 + A)
= cosec(π/2 + A) = -secA

sec(3π + A) = sec(2π + π + A)
= sec(π + A) = -secA

cot(A - π/2) = -cot(π/2 - A) = -tanA

now, LHS = {sin(3π - A)cos(A - π/2)tan(3π/2 - A)}{cosec(13π/2 + A)sec(3π + A)cot(A - π/2)}

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

= {sin²A cotA}/{sec²A tanA}

= {sin²A × cosA/sinA}/{1/cos²A × sinA/cosA}

= cosA/{1/cos³A}

= cos⁴A = RHS

Answer 2

HELLO DEAR,



we know:-
sin(3π - A) = sin{2π + ( π - A)}
= sin(π - A) = sinA
cos(A - π/2) = cos{-(π/2 - A)} = cos(π/2 - A)
= cos(π/2 - A) = sinA
tan(3π/2 - A) = cotA
cosec(13π/2 + A) = cosec(6π + π/2 + A)
= cosec(π/2 + A) = -secA
sec(3π + A) = sec(2π + π + A)
= sec(π + A) = -secA
cot(A - π/2) = -cot(π/2 - A) = -tanA

now,
{sin(3π - A)cos(A - π/2)tan(3π/2 - A)}{cosec(13π/2 + A)sec(3π + A)cot(A - π/2)}

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

=> {sin²A cotA}/{sec²A tanA}

=> {sin²A × cosA/sinA}/{1/cos²A × sinA/cosA}

=> cosA/{1/cos³A}

=> cos⁴A



I HOPE IT'S HELP YOU DEAR,
THANKS