Question

p.sin3a+q.cosa=sina.cosa ,p.sina-q.cosa ,so proof that P2+q2=1​

Answer 1

Answer:

We have:

tanA+sinA=p=>(tanA+sinA)2=p2=>

p2=tan2A+sin2A+2tanAsinA…(1)

Now, if tanA−sinA=q , then we take:

tanA−sinA=q=>(tanA−sinA)2=q2=>

q2=tan2A+sin2A−2tanAsinA…(2)

By (1) , (2) , it follows that:

p2−q2=4(tanA)(sinA)…(3)

Now, we are also given that:

p2−q2=4pq−−√=>p2−q2=4sqrttan2A−sin2A=>

p2−q2=4(sin2A/cos2A)−sin2A−−−−−−−−−−−−−−−−−−−√=>

p2−q2=4[sin2A−(cos2A)sin2A]/cos2A−−−−−−−−−−−−−−−−−−−−−−−−√=>

p2−q2=4[sin2A(1−cos2A)]/cos2A−−−−−−−−−−−−−−−−−−−−−√=>

p2−q2=4[sin2A(sin2A)]/cos2A−−−−−−−−−−−−−−−−−−√=>

p2−q2=4[(sin2A)/cosA]=>p2−q2=4(tanA)(sinA)…(4)

By (3) and (4) , we conclude that if tanA+sinA=p and tanA−sinA=q , then:

p2−q2=4pq−−√

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