CBSE Class 11-science Answered
If SinA=2pq/p[square}+q[square]
Pt SecA-Tan A=p-q/p+q
Asked by chethan U | 22 Jul, 2012, 05:27: PM
Expert Answer
Answer : Given :SinA=2pq / (p2+q2)
To Prove : SecA - Tan A= (p - q ) / (p+q)
Now ,
=> SinA = 2pq / (p2+q2)
Using componendo and dividendo,we get
=> ( sinA +1 ) / (sinA -1) = ( 2pq + p2 + q2 ) / (p2 + q2 - 2pq )
Inverting the values we get ,
=> ( sinA - 1 ) / (sinA + 1) = (p2 + q2 - 2pq ) / ( 2pq + p2 + q2)
=> ( sinA - 1 ) / (sinA + 1) = (p - q)2 / ( p + q)2
rationalising LHS
=> (1- sinA)2 / ( 1- sin2A ) = (p - q)2 / ( p + q)2
=> (1- sinA)2 / cos2A = (p - q)2 / ( p + q)2 { using sin2A + cos2A =1 ,
=> 1- sin2A = cos2A }
taking the squareroot both the sides , we get
=> (1- sinA) / cosA = (p - q) / ( p + q)
=>(1/ cosA) - (sinA / cosA) = (p - q) / ( p + q )
=> secA - tanA = (p - q) / ( p + q) { using secA = 1/cosA and tanA = sinA/cosA }
Hence Proved
Answered by | 22 Jul, 2012, 08:11: PM
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