Question

If p = (cot A + tan A) and
q = (sec A + cosec A), prove that
(92 - p2)2 = 4p2​

Answer 1

Step-by-step explanation:

#include<stdio.h>

void main()

{

int arr[5]={4,12,78,90,70};

int i=0;

for(i=0;i<=4;i++)

{

printf("\n %d",arr[i]);

}

return 0;

}

Answer 2

Answer:

i think some thing i missed here !!

Step-by-step explanation:

p = (cot A + tan A) & q = (sec A + cosec A)

to prove: (p²-q²)² = 4p²  ---------------- (A)

=> (p²-q²) = √(4p²) = 2p

ie., to prove (A) , we just need to prove  (p²-q²) = 2p

let us take the

LHS = (p²-q²)

        = (cot A + tan A)² - (sec A + cosec A)²        

        = (cot²A +2+tan²A) - (sec²A +2secA.cosecA+cosec²A)

        = (cot²A +1+1+tan²A) - (sec²A +2secA.cosecA+cosec²A)

        = cosec²A+sec²A - sec²A - 2 secA.cosecA - cosec²A

        = - 2 secA.cosecA

       

        = - 2($\frac{1}{cosA}$*$\frac{1}{sinA}$)

        = 2* (  $\frac{1}{ ( \frac{sin2A}{2} ) }$  )

       

2p = 2(cotA+tanA) = 2(1/tanA  + tanA )

     = 2(1+tan²A)/tanA

     = 2 sec²A . cotA