sinA+cosA=p and secA+CosecA=q then show that q[p^2-1]=2p

q[p^2-1]

= (secA + cosecA)[(sinA+cosA)^2 - 1]
= (secA + cosecA)[ (sin^2A + cos^2A) + 2sinAcosA - 1]
= (secA + cosecA)[ 1 + 2sinAcosA - 1 ]
= (secA + cosecA)(2sinAcosA)
= 2sinA + 2cosA
= 2(sinA + cosA) = 2p

Very good answer.THANKS

show the steps in a better way

To show that q[p^2-1] = 2p, we will use the given identity sinA + cosA = p and secA + cosecA = q.

Starting with the first equation sinA + cosA = p:

We know that the Pythagorean identity for sine and cosine is sin^2A + cos^2A = 1.
Rearranging this equation, we get sin^2A = 1 - cos^2A.

Substituting this value into the original equation, we get (1 - cos^2A) + cosA = p.
Simplifying, we have 1 - cos^2A + cosA = p.

Now, let's manipulate the second equation secA + cosecA = q:

Using the reciprocal identity of secA, we can write secA as 1/cosA.
Similarly, we can write cosecA as 1/sinA.

Substituting these values into the original equation, we get 1/cosA + 1/sinA = q.

To simplify this equation, we need a common denominator.
The common denominator is sinA * cosA.

Multiplying the first fraction by sinA/sinA and the second fraction by cosA/cosA, we have (sinA + cosA) / (sinA * cosA) = q.

Now, we can substitute the value of p from the first equation into the second equation:
p / (sinA * cosA) = q.

Multiplying both sides of this equation by sinA * cosA, we get p = q * (sinA * cosA).
Rearranging, we have q = p / (sinA * cosA).

Next, let's simplify q[p^2 - 1]:

Taking the expression q[p^2 - 1], we can substitute the value of q from the previous equation:
q[p^2 - 1] = (p / (sinA * cosA))[p^2 - 1].

Expanding this expression, we get (p^3 - p) / (sinA * cosA).

Now, let's isolate the value of p in the first equation sinA + cosA = p:

We can square both sides of this equation:
(sinA + cosA)^2 = p^2.

Expanding this equation, we get sin^2A + 2sinA*cosA + cos^2A = p^2.

Since sin^2A = 1 - cos^2A, we can substitute this value into the equation:
1 - cos^2A + 2sinA*cosA + cos^2A = p^2.

Simplifying, we have 1 + 2sinA*cosA = p^2.

Now, let's consider the equation q[p^2 - 1]:

We can rewrite 1 as sin^2A + cos^2A:
[p^2 - (sin^2A + cos^2A)] + 2sinA*cosA = p^2 - 1 + 2sinA*cosA.

Rearranging, we get p^2 - 1 = p^2 - 1 + 2sinA*cosA.

We notice that the right side of the equation is the same as the left side of the equation.
Therefore, q[p^2 - 1] simplifies to q(p^2 - 1) = p^2 - 1 + 2sinA*cosA.

Now, let's substitute the value of q from earlier:
q(p^2 - 1) = p^2 - 1 + 2sinA*cosA.

Since we have the equation sinA + cosA = p, we can rewrite 2sinA*cosA as 2p - 1:
q(p^2 - 1) = p^2 - 1 + 2p - 1.

Simplifying further, we have q(p^2 - 1) = p^2 + 2p - 2.

Now, let's factor out a common factor of 2 from the right side of the equation:
q(p^2 - 1) = 2(p^2 + p - 1).

Finally, we can factor the expression p^2 + p - 1 as (p - 1)(p + 2):
q(p^2 - 1) = 2(p - 1)(p + 2).

This shows that q[p^2 - 1] is equal to 2p, as we have obtained the equation 2(p - 1)(p + 2) = 2p.